Question:

A ball thrown up from the ground reaches a maximum height of 20 m. Find

• (a) Its initial velocity
• (b) The time taken to reach the highest point
• (c) Its velocity just before hitting the ground
• (d) its displacement between 0.5 s and 2.5 s
• (e) The time at which it is 15 m above the ground

Solution:

(a) Its initial velocity

$${u=\sqrt{2gh}}$$

$${u=\sqrt {(2)(10)(20)}}$$

$${u=\sqrt 400 =20 \frac{m}{s}}$$

Its initial velocity is 20 ms^-1

(b). The time taken to reach the maximum height

$${u=gt}$$

$${t=\frac{u}{g}}$$

$${t=\frac {20}{10}}$$

$${t=2 s}$$

The time taken to reach maximum height is 2 sec.

(c).Its velocity just before hitting the ground

$${v=\sqrt{2gh}}$$

$${v=\sqrt{(2)(10)(20)}}$$

$${v=\sqrt{400}}$$

$${v=20\frac{m}{s}}$$

Its velocity just before hitting the ground is 20 ms^-1

(d). Its displacement between 0.5 s and 2.5 s

Displacement at 0.5 s

$$$$

$${s=(20)(0.5)-\frac{1}{2}(10)(0.5^2)}$$

$${s=10-5(0.25)}$$

$${s=8.75 m}$$

Displacement at 2.5 s

$${s=(20)(2.5)-\frac{1}{2}(10)(2.5^2)}$$

$${s=50-(5)(6.25)}$$

$${s=18.75 m}$$

Its displacement between 0.5 s and 2.5 s is=18.75-8.75 = 10 m

(e) The time at which it is 15 m above the ground

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

$${15=20t-\frac{1}{2}(10)t^2}$$

$${5t^2-20t+15}=0$$

$${t^2-4t+3}=0$$

$${t= 1s , t=3s}$$