1. Home
  2. Kinematics fundamentals
  3. Accelerated motion in two
  4. Projectile motion
  5. Projectile motion on an incline

1.6 Projectile motion on an incline  (Page 3/7)

<< Chapter < Page
  Kinematics fundamentals     Page 3 / 7
Chapter >> Page >

Range of flight

First thing that we should note that we do not call “horizontal range” as the range on the incline is no more horizontal. Rather we simply refer the displacement along x-axis as “range”. We can find range of flight by considering motion in both “x” and “y” directions. Note also that we needed the same approach even in the normal case. Let “R” be the range of projectile motion.

The motion along x-axis is no more uniform, but decelerated. This is the major difference with respect to normal case.

x = u x T − 1 2 a x T 2

Substituting value of “T” as obtained before, we have :

R = u cos θ − α X 2 u sin θ − α g cos α − g sin α X 4 u 2 sin 2 θ − α 2 g 2 cos 2 α

⇒ R = u 2 g cos 2 α { 2 cos θ − α sin θ − α cos α − sin α X 2 sin 2 θ − α }

Using trigonometric relation, 2 sin 2 θ − α = 1 − cos 2 θ − α ,

⇒ R = u 2 g cos 2 α [ sin 2 θ − α cos α − sin α { 1 − cos 2 θ − α } ]

⇒ R = u 2 g cos 2 α { sin 2 θ − α cos α − sin α + sin α cos 2 θ − α }

We use the trigonometric relation, sin A + B = sin A cos B + cos A sin B ,

⇒ R = u 2 g cos 2 α { sin 2 θ − 2 α + α − sin α }

⇒ R = u 2 g cos 2 α { sin 2 θ − α − sin α }

This is the expression for the range of projectile on an incline. We can see that this expression reduces to the one for the normal case, when α = 0 ,

⇒ R = u 2 sin 2 θ g

Maximum range

The range of a projectile thrown up the incline is given as :

R = u 2 g cos 2 α { sin 2 θ − α − sin α }

We see here that the angle of incline is constant. The range, therefore, is maximum for maximum value of “sin(2θ – α)”. Thus, range is maximum for the angle of projection as measured from horizontal direction, when :

sin 2 θ − α = 1

⇒ sin 2 θ − α = sin π / 2

⇒ 2 θ − α = π / 2

⇒ θ = π / 2 + α / 2

The maximum range, therefore, is :

⇒ R max = u 2 g cos 2 α 1 − sin α

Problem : Two projectiles are thrown with same speed, “u”, but at different angles from the base of an incline surface of angle “α”. The angle of projection with the horizontal is “θ” for one of the projectiles. If two projectiles reach the same point on incline, then determine the ratio of times of flights for the two projectiles.

Projectile motion up an incline

Two projectiles reach the same point on the incline.

Solution : We need to find the ratio of times of flights. Let T 1 and T 2 be the times of fights. Now, the time of flight is given by :

T = 2 u sin θ − α g cos α

Here, the angle of projection of one of the projectiles, “θ”, is given. However, angle of projection of other projectile is not given. Let “θ’” be the angle of projection of second projectile.

⇒ T 1 T 2 = 2 u sin θ − α 2 u sin θ ′ − α

We need to know “θ’” to evaluate the above expression. For this, we shall make use of the fact that projectiles have same range for two angles of projections. We can verify this by having a look at the expression of range, which is given as :

⇒ R = u 2 g cos 2 α { sin 2 θ − α − sin α }

Since other factors remain same, we need to analyze motions of two projectiles for same range in terms of angle of projection only. We have noted in the case of normal projectile motion that there are complimentary angle for which horizontal range is same. Following the same line of argument and making use of the trigonometric relation sinθ = sin (π -θ), we analyze the projectile motions of equal range. Here,

sin 2 θ ′ − α = sin { π − 2 θ − α } = sin π − 2 θ + α

2 θ ′ − α = π − 2 θ + α

⇒ 2 θ ′ = π − 2 θ + 2 α

⇒ θ ′ = π 2 − θ + α

Putting this value in the expression for the ratio of times of flights, we have :

⇒ T 1 T 2 = 2 u sin θ − α 2 u sin π / 2 − θ + α − α

⇒ T 1 T 2 = sin θ − α sin π / 2 − θ

⇒ T 1 T 2 = sin θ − α cos θ

Got questions? Get instant answers now!
<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Kinematics fundamentals. OpenStax CNX. Sep 28, 2008 Download for free at http://cnx.org/content/col10348/1.29
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Kinematics fundamentals' conversation and receive update notifications?

Ask