Question

2. A ball rolls off the edge of a table 1.44m above the floor and strikes the floor at a point 2m horizontally from the edge of the table. (5 pts)
A) What is the time the ball was in the air?

B) What is the initial velocity of the ball?

Answer 1

Explanation:

The initial velocity of the ball was +4.90 m/s (up).

Answer 2

A) The time the ball was in the air is  $0.54s$

B)The initial velocity of the ball is $3.7m/s$

Step by step explanation

Given:

Height of the table$=1.44m$

horizontal distance $=2m$

To find:

• The time the ball was in the air

• Initial velocity of the ball

Formula used:

• $h=\frac{1}{2}$×$g$×$t^{2}$

where: $h=$height(m)

$g=9.8m/s$

$t=$time(s)

• $D=$$v$×$t$

where:$D=$horizontal distance

$v=$velocity

$t=$time

Using the equation $h=\frac{1}{2}$×$g$×$t^{2}$ we can find the time the ball was in the air.

$h=\frac{1}{2}$×$g$×$t^{2}$

$1.44=\frac{1}{2}$×$9.8$×$t^{2}$

$t^{2} =\frac{2(1.44)}{9.8}$

$t^{2} =\frac{2.88}{9.8}$

$t=\sqrt{0.293877551}$

$t=0.54s$

using this value of time we can calculate the initial velocity using the equation:

$D=v$×$t$

$2=v$×$0.54$

$v=\frac{2}{0.54}$

$v=3.7m/s$

Therefore for the ball rolling off the edge of table of height $1.44m$, strikes the floor at a point $2m$ and the time taken for the ball to be in the air is$0.54s$  and the initial velocity of the ball is found to be $3.7m/s$.