Question

What was the direction of the ball’s velocity

Answer 1

Part of the question is missing. Here it is:

A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table.     The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity  just before it hit the floor?

Answer:

$\theta=-75.7^{\circ}$

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

$s=ut+\frac{1}{2}at^2$

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

t is the time

Solving for t,

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s$

Now we can find the final vertical velocity of the ball, using:

$v_y=u+at$

And susbtituting t = 0.52 s, we find

$v_y = 0 +(9.8)(0.52)=5.1 m/s$

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

$v_x = \frac{0.7}{0.52}=1.3 m/s$

So now we can find the direction of the ball's velocity using:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}$

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

$\theta=-75.7^{\circ}$