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

Answer:

Initial velocity = 39.2m/s

Maximum height is 78.4m

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

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

This becomes

$s = ut - \frac{1}{2}gt^2$

Substitute values for u, t and g

$s = 39.2 * 4 - \frac{1}{2} * 9.8 * 4^2$

$s = 156.8 - 78.4$

$s = 78.4$

Hence, the maximum height is 78.4m