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}$$
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