Question

Simple question please help! A ball is thrown 19.0 m/s at an angle of 40.0º with the horizontal. Assume the ball is thrown at ground level 22.0 meters away. a. How long does it take the ball to reach the wall? b. At what height does the ball hit the wall?

Answer 1

a. 1.51 s

In this part of the problem, we are only interested in the horizontal motion of the ball. Along the horizontal direction, the motion of the ball is a uniform motion with constant velocity, which is equal to the horizontal component of the initial velocity:

$v_x = v_0 cos \theta = (19.0 m/s)(cos 40^{\circ})=14.6 m/s$

So, the ball travels horizontally at a speed of 14.6 m/s; in order to cover the distance of d = 22.0 m that separates it from the wall, the time need is

$t=\frac{d}{v_x}=\frac{22.0 m}{14.6 m/s}=1.51 s$

b. 7.25 m

We now know that the ball takes 1.51 s to hit the wall, 22.0 away. Now we have to analyze the vertical motion of the ball, which is an accelerated motion with constant acceleration g =9.8 m/s^2 towards the ground (acceleration due to gravity).

The initial vertical velocity is

$v_{y0}=v_0 sin \theta=(19.0 m/s)(sin 40^{\circ})=12.2 m/s$

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

$y(t)=v_{0y} t -\frac{1}{2}gt^2$

We know that the ball hits the wall at t=1.51 s, so if we substitute this value into the previous formula, we find at what height y the ball hits the wall:

$y(1.51 s)=(12.2 m/s)(1.51 s)-\frac{1}{2}(9.8 m/s^2)(1.51 s)^2=7.25 m$