Question

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She throws the softball with a velocity of 21.5 m/s at an angle of 33.5∘ above the horizontal. When the softball leaves her hand, it is 13.5 m above the water. How far does the softball travel horizontally before it hits the water?

Answer 1

Answer:

58.5 m

Explanation:

First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.

The initial vertical velocity of the ball is

$u_y = u sin \theta$

where

u = 21.5 m/s is the initial speed

$\theta=33.5^{\circ}$ is the angle

Substituting,

$u_y = (21.5) sin 33.5^{\circ} =11.9 m/s$

The vertical position of the ball at time t is given by

$y = h + u_y t + \frac{1}{2}gt^2$

where

h = 13.5 m is the initial heigth

$g = -9.8 m/s^2$ is the acceleration of gravity (negative sign because it points downward)

The ball reaches the water when y = 0, so

$0 = h + u_yt +\frac{1}{2}gt^2\\0 = 13.5 +11.9 t - 4.9t^2$

Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.

The horizontal velocity of the ball is

$u_y = u cos \theta = (21.5) cos 33.5^{\circ} =17.9 m/s$

And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:

$d= u_x t = (17.9)(3.27)=58.5 m$