Chapter 3, Problem 88GP

A projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high (see Fig. 3-60). If the projectile lands on top of the cliff 6.6 s after it is fired, find the initial velocity of the projectile (magnitude and direction). Neglect air resistance.

FIGURE 3-60

Problem 88.

88. Choose the origin to be the point from which the projectile is launched, and choose upward as the positive y direction. The y displacement of the projectile is 135 m, and the horizontal range of the projectile is 195 m. The acceleration in the y direction is a_y = −g, and the time of flight is 6.6 s.

The horizontal velocity is found from the horizontal motion at constant velocity.

$\Delta x=v_{x}t\to v_x=\frac{\Delta x}{t}=\frac{195\text{m}}{6.6\text{s}}=29.55\text{ m/s}$

Calculate the initial v velocity from the given data and Eq. 2-12b.

$y=y_0+v_{y0}t+\frac{1}{2}{a}_y{t}^2\to 135\text{m}=v_{y0}(6.6\text{s})+\frac{1}{2}{(-9.80\text{ m/s)}}^{\text{2}}\text{ = 60 m/s}\to v_{y0}=52.79\text{ m/s}$

Thus the initial velocity and direction of the projectile are as follows.

$\begin{array}{c}{v}_0=\sqrt{v_x^2+v_{y0}^2}=\sqrt{{(29.55\text{ m/s)}}^{\text{2}}\text{ + (52}{\text{.79 m/s)}}^{\text{2}}}=\text{ 60m/s}\\ \theta ={\tan }^{-1}\frac{v_{y0}}{v_x}={\tan }^{-1}\frac{52.79\text{ m/s}}{29.55\text{ m/s}}=61°\end{array}$