Respuesta :
Answer:
1. 2 second
2. 7.83 second
3. 58.31 m/s
Explanation:
initial velocity, u = 20 m/s
g = 10 m/s^2
1. Let it takes time t1 to reach to maximum height.
At maximum height the final velocity of the ball is zero.
use first equation of motion, we get
v = u + at
0 = 20 - 10 t1
t1 = 2 second
Thus, the time taken to reach to maximum height is 2 second.
2. The maximum height above the cliff is h
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
0 = 20 x 20 - 2 x 10 x h
400 = 20 h
h = 20 m
The total height is 20 + 150 = 170 m
let the time taken by the ball to reach to bottom from maximum height is t2.
use third equation of motion
[tex]s = ut + \frac{1}{2}at^{2}[/tex]
170 = 0 + 0.5 x 10 x t2^2
t2 = 5.83 second
thus, the total time to reach to bottom, t = t1 + t2 = 2 + 5.83 = 7.83 second.
3. Let v be the velocity with which the ball strikes the ground.
Use third equation of motion, we get
[tex]v^{2}=u^{2}+2as[/tex]
[tex]v^{2}=0^{2}+2\times 10 \times 170[/tex]
v = 58.31 m/s
Thus, the ball reaches the ground with velocity of 58.31 m/s.
