Name: Understanding Instantaneous Speed
Uploaded: 2019-07-19
Duration: 5 min 28 s
Description: How fast is a ball just after a bowler releases it from his hand? How fast is it when it reaches the target? In this video, we discuss speed at any point on...

Question

A steel ball is dropped from a building's roof and passes a window, taking

0.125 s

to fall a distance of

1.20 m

from the top to the bottom of the window. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in

0.125 s

. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is

2.00 s

. How tall is the building?

Accepted Answer

Given:

The time taken from $A$ to $B$ is $0.125 sec$ .

Distance from $A$ to $2.00 s$ is $1.2 m$ .

Time from $B$ to $C$ and again to $B$ is $2 sec$ , so time taken from $B$ to $C$ will be $1 sec$ .

From $B$ to $C$ , the acceleration is constant, so the average velocity can be given by

$v_{av} = \frac{v_{A} + v_{B}}{2} = \frac{s_{AB}}{t_{AB}} = \frac{1.2}{0.125}$ ,

where $s_{AB}$ is displacement and $t_{AB}$ is time taken from $A$ to $B$ .

Solving the above equation,

$v_{A} + v_{B} = 9.6 . . . (1)$

Again from $A$ to $B$ , applying the equation of motion,

$v_{B} = v_{A} + g \times t_{AB}$

$v_{B} - v_{A} = 9.8 \times 0.125$

$v_{B} - v_{A} = 1.225 . . . (2)$

Solving equations $(1)$ and $(2)$ ,

$v_{A} = 5.41 {m s}^{- 1}$ and

$v_{B} = 6.63 {m s}^{- 1}$

Let the distance from $P$ to $A$ be $h_{1}$ . We then apply the equation of motion from $P$ to $A$ .

${(v_{A})}^{2} = {(u)}^{2} + 2 g h_{1}$

${(5.41)}^{2} = {(0)}^{2} + 2 \times 9.8 \times h_{1}$

$h_{1} = 1.49$

If the distance from $B$ to $C$ is taken as $h_{2}$ , then,

again applying the equation of motion from $B$ to $C$ ,

$s_{BC} = v_{B} \times t_{BC} + \frac{1}{2} g \times {(t_{BC})}^{2}$

$h_{2} = 6.63 \times 1 + \frac{1}{2} \times 9.8 \times {(1)}^{2}$

$h_{2} = 1.53$

The total height of the building is

$H = h_{1} + 1.2 + h_{2}$

$H = 1.49 + 1.2 + 1.53$

$H = 4.22 m$ .

Important Questions on Motion Along a Straight Line

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