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

a) 9.8 m/s

b) 4.9 m

Step-by-step explanation:

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

`y=y_(o)+V_(o)t-(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-(1)/(2)gt^(2)` (3)

Isolating
`V_(o)`:

`V_(o)=(1)/(2)gt` (4)

`V_(o)=(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`