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0.4 2.5 projectile motion  (Page 6/13)

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R = v 0 2 sin 2 θ 0 g , size 12{R= { {v rSub { size 8{0} } rSup { size 8{2} } "sin"2θ rSub { size 8{0} } } over {g} } ","} {}

where v 0 size 12{v rSub { size 8{0} } } {} is the initial speed and θ 0 size 12{θ rSub { size 8{0} } } {} is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.

Summary

  • Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
  • To solve projectile motion problems, perform the following steps:
    1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s size 12{s} {} are given by the quantities x size 12{x} {} and y size 12{y} {} , and the components of the velocity v size 12{v} {} are given by v x = v cos θ size 12{v rSub { size 8{x} } =v"cos"θ} {} and v y = v sin θ size 12{v rSub { size 8{y} } =v"sin"θ} {} , where v size 12{v} {} is the magnitude of the velocity and θ size 12{θ} {} is its direction.
    2. Analyze the motion of the projectile in the horizontal direction using the following equations:
      Horizontal motion ( a x = 0 ) size 12{"Horizontal motion " \( a rSub { size 8{x} } =0 \) } {}
      x = x 0 + v x t size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {}
      v x = v 0 x = v x = velocity is a constant. size 12{v rSub { size 8{x} } =v rSub { size 8{0x} } =v rSub { size 8{x} } ="velocity is a constant."} {}
    3. Analyze the motion of the projectile in the vertical direction using the following equations:
      Vertical motion ( Assuming positive direction is up; a y = − g = − 9 . 80 m /s 2 ) size 12{"Vertical motion " \( "Assuming positive direction is up; "a rSub { size 8{y} } = - g= - 9 "." "80"" m/s" rSup { size 8{2} } \) } {}
      y = y 0 + 1 2 ( v 0 y + v y ) t size 12{y=y rSub { size 8{0} } + { {1} over {2} } \( v rSub { size 8{0y} } +v rSub { size 8{y} } \) t} {}
      v y = v 0 y − gt size 12{v rSub { size 8{y} } =v rSub { size 8{0y} } - ital "gt"} {}
      y = y 0 + v 0 y t − 1 2 gt 2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {}
      v y 2 = v 0 y 2 − 2 g ( y − y 0 ) . size 12{v rSub { size 8{y} } rSup { size 8{2} } =v rSub { size 8{0y} } rSup { size 8{2} } - 2g \( y - y rSub { size 8{0} } \) } {}
    4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:
      s = x 2 + y 2 size 12{s= sqrt {x rSup { size 8{2} } +y rSup { size 8{2} } } } {}
      θ = tan − 1 ( y / x ) size 12{θ="tan" rSup { size 8{ - 1} } \( y/x \) } {}
      v = v x 2 + v y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {}
      θ v = tan − 1 ( v y / v x ) . size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } \( v rSub { size 8{y} } /v rSub { size 8{x} } \) } {}
  • The maximum height h size 12{h} {} of a projectile launched with initial vertical velocity v 0 y size 12{v rSub { size 8{0y} } } {} is given by
    h = v 0 y 2 2 g . size 12{h= { {v rSub { size 8{0y} } rSup { size 8{2} } } over {2g} } } {}
  • The maximum horizontal distance traveled by a projectile is called the range . The range R size 12{R} {} of a projectile on level ground launched at an angle θ 0 size 12{θ rSub { size 8{0} } } {} above the horizontal with initial speed v 0 size 12{v rSub { size 8{0} } } {} is given by
    R = v 0 2 sin 2 θ 0 g . size 12{R= { {v rSub { size 8{0} } rSup { size 8{2} } "sin"2θ rSub { size 8{0} } } over {g} } } {}

Conceptual questions

Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º size 12{"0°"} {} nor 90º size 12{"90°"} {} ): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at t = 0 size 12{t=0} {} ? (d) Can the speed ever be the same as the initial speed at a time other than at t = 0 size 12{t=0} {} ?

Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º size 12{"0°"} {} nor 90º size 12{"90°"} {} ): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite in direction to a component of velocity?

Problems&Exercises

A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x size 12{x} {} and y size 12{y} {} distances from where the projectile was launched to where it lands?

x = 1.30 m × 10 2 y = 30 .9 m.

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

(a) 3.50 s

(b) 28.6 m/s (c) 34.3 m/s

(d) 44.7 m/s, 50.2º below horizontal

The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6 . 37 × 10 3 km size 12{6 "." "37" times "10" rSup { size 8{3} } " km"} {} . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

(a) 560 m/s

(b) 8 . 00 × 10 3 m size 12{8 "." "00 " times " 10" rSup { size 8{3} } " m"} {}

(c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

An arrow is shot from a height of 1.5 m toward a cliff of height H size 12{H} {} . It is shot with a velocity of 30 m/s at an angle of 60º size 12{"60°"} {} above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ size 12{θ} {} below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle θ size 12{θ} {} such that the ball just crosses the net? Will the ball land in the service box, whose out line is 6.40 m from the net?

θ = 6.1º size 12{θ} {}

yes, the ball lands at 5.3 m from the net

An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0º size 12{"30º} below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

4.23 m. No, the owl is not lucky; he misses the nest.

Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

No, the maximum range (neglecting air resistance) is about 92 m.

The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

A basketball player is running at 5 . 00 m/s size 12{5 "." "00 m/s"} {} directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.750 m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?

A football player punts the ball at a 45.0º size 12{"45"°} {} angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?

(a) 24.2 m/s

(b) The ball travels a total of 57.4 m with the brief gust of wind.

Questions & Answers

what does the ideal gas law states
Joy Reply
Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
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Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
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8m/s²
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which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
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I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
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Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
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about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
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physics, biology and chemistry this is my Field
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field is a region of space under the influence of some physical properties
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Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
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like charges repel while unlike charges atttact
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Source:  OpenStax, 2d kinematics. OpenStax CNX. Sep 04, 2015 Download for free at http://legacy.cnx.org/content/col11879/1.3
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