4.3 Projectile motion  (Page 6/13)

Substituting the expression for t into the equation for the position $y=(v_0\text{sin}{\theta }_0)t-\frac{1}{2}g{t}^2$ gives

Rearranging terms, we have

This trajectory equation is of the form $y=ax+b{x}^2,$ which is an equation of a parabola with coefficients

Range

From the trajectory equation we can also find the range    , or the horizontal distance traveled by the projectile. Factoring [link] , we have

The position y is zero for both the launch point and the impact point, since we are again considering only a flat horizontal surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch point, and

corresponding to the impact point. Using the trigonometric identity $2\text{sin}\theta \text{cos}\theta =\text{sin}2\theta$ and setting x = R for range, we find

Note particularly that [link] is valid only for launch and impact on a horizontal surface. We see the range is directly proportional to the square of the initial speed $v_0$ and $\text{sin}2{\theta }_0$ , and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, we see from the factor $\text{sin}2{\theta }_0$ that the range is maximum at $45\text{°}.$ These results are shown in [link] . In (a) we see that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at $45\text{°}.$ This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is found for two initial launch angles that sum to $90\text{°}.$ The projectile launched with the smaller angle has a lower apex than the higher angle, but they both have the same range.

Comparing golf shots

A golfer finds himself in two different situations on different holes. On the second hole he is 120 m from the green and wants to hit the ball 90 m and let it run onto the green. He angles the shot low to the ground at $30\text{°}$ to the horizontal to let the ball roll after impact. On the fourth hole he is 90 m from the green and wants to let the ball drop with a minimum amount of rolling after impact. Here, he angles the shot at $70\text{°}$ to the horizontal to minimize rolling after impact. Both shots are hit and impacted on a level surface.

(a) What is the initial speed of the ball at the second hole?

(b) What is the initial speed of the ball at the fourth hole?

(c) Write the trajectory equation for both cases.

(d) Graph the trajectories.

Strategy

We see that the range equation has the initial speed and angle, so we can solve for the initial speed for both (a) and (b). When we have the initial speed, we can use this value to write the trajectory equation.

Solution

(a) $R=\frac{v_0^2\text{sin}2{\theta }_0}{g}\Rightarrow v_0=\sqrt{\frac{Rg}{\text{sin}2{\theta }_0}}=\sqrt{\frac{90.0\text{m}(9.8\text{m}\text{/}{\text{s}}^2)}{\text{sin}(2(70\text{°}))}}=37.0\text{m}\text{/}\text{s}$

(b) $R=\frac{v_0^2\text{sin}2{\theta }_0}{g}\Rightarrow v_0=\sqrt{\frac{Rg}{\text{sin}2{\theta }_0}}=\sqrt{\frac{90.0\text{m}(9.8\text{m}\text{/}{\text{s}}^2)}{\text{sin}(2(30\text{°}))}}=31.9\text{m}\text{/}\text{s}$

(c)
$\begin{array}{}\\ \\ \\ y=x\left[\text{tan}{\theta }_0-\frac{g}{2{(v_0\text{cos}{\theta }_0)}^2}x\right]\hfill \\ \text{Second hole:}y=x\left[\text{tan}70\text{°}-\frac{9.8\text{m}\text{/}{\text{s}}^2}{2{[(37.0\text{m}\text{/}\text{s)(}\text{cos}70\text{°})]}^2}x\right]=2.75x-0.0306x^2\hfill \\ \text{Fourth hole:}y=x\left[\text{tan}30\text{°}-\frac{9.8\text{m}\text{/}{\text{s}}^2}{2{[(31.9\text{m}\text{/}\text{s)(}\text{cos}30\text{°})]}^2}x\right]=0.58x-0.0064x^2\hfill \end{array}$

(d) Using a graphing utility, we can compare the two trajectories, which are shown in [link] .

Significance

The initial speed for the shot at $70\text{°}$ is greater than the initial speed of the shot at $30\text{°}.$ Note from [link] that two projectiles launched at the same speed but at different angles have the same range if the launch angles add to $90\text{°}.$ The launch angles in this example add to give a number greater than $90\text{°}.$ Thus, the shot at $70\text{°}$ has to have a greater launch speed to reach 90 m, otherwise it would land at a shorter distance.

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