Question

A ball thrown vertically upward is caught by the thrower after 4.00 sec. Find the initial velocity of the ball and the maximum height it reaches.

Answer 1

Initial velocity = 39.2m/s

Maximum height is 78.4m

Step-by-step explanation:

Given

`Time, t = 4s`

Solving (a): Initial Velocity

Using first law of motion:

`v = u + at`

Where

`v = final\ velocity = 0`

`u = iniital\ velocity = ??`

`a = acceleration = -g` [g represents acceleration due to gravity]

`t = 4`

Substitute these value in the above formula:

`v = u + at`

`0 = u - g * 4`

`0 = u - 9.8 * 4`

Take g as 9.8m/s²

`0 = u - 39.2`

`u = 39.2m/s\\`

Hence, initial velocity = 39.2m/s

Solving (b): Maximum Height

This will be solved using second equation of motion

`s = ut + (1)/(2)at^2`

This becomes

`s = ut - (1)/(2)gt^2`

Substitute values for u, t and g

`s = 39.2 * 4 - (1)/(2) * 9.8 * 4^2`

`s = 156.8 - 78.4`

`s = 78.4`

Hence, the maximum height is 78.4m