Question

A ball thrown vertically upward is caught by the thrower after 2.00 s. Find (a) the initial velocity of the ball and (b) the maximum height the ball reaches.

Answer 1

Answer:

a)  9.8 m/s

b) 4.9 m

Explanation:

This problem is a good example of Vertical motion, where the main equations for this situation are:  

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)  

$V^{2}={V_{o}}^{2}-2gy$ (2)  

Where:  

$y$ is the height of the ball at a given time

$y_{o}=0m$ is the initial height of the ball (assuming the hand of the thrower the origin of the system)  

$V_{o}$ is the initial velocity of the ball

$V$ is the final velocity of the ball

$t=2s$ is the time it takes for the ball to make the complete movement (from the moment it is thrown until it falls back into the pitcher's hands)

$g=9.8 m/s^{2}$ is the acceleration due to gravity  

Knowing this, let's begin with the answers:

a) Initial velocity

In order to find the initial velocity $V_{o}$ of the ball, we will use equation (1) and $t=2s$, taking into account that $y=0 m$ and $y_{o}=0m$ at this given time:

$0=0+V_{o}t-\frac{1}{2}gt^{2}$ (3)  

Isolating $V_{o}$:

$V_{o}=\frac{1}{2}gt$ (4)  

$V_{o}=\frac{1}{2}(9.8 m/s^{2})(2 s)$ (5)  

Then:

$V_{o}=9.8 m/s$ (6)  

b) Maximum height

In this part, we will use equation (2), knowing the value of the height is maximum when $V=0$. So, we will name this height as $y_{max}$:

$0={V_{o}}^{2}-2gy_{max}$ (7)  

Isolating $y_{max}$:

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8)  

$y_{max}=\frac{{(9.8 m/s)}^{2}}{2(9.8 m/s^{2})}$ (9)  

Finally:

$y_{max}=4.9 m$