Question

A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.How much later does it reach the bottom of the cliff?What is its speed just before hitting?What total distance did it travel?

Answer 1

5.72 seconds

848.27 m/s

97.94 m

Step-by-step explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

`v=u+at\\\Rightarrow 0=15-9.81* t\\\Rightarrow (-15)/(-9.81)=t\\\Rightarrow t=1.52 \s`

Time taken to reach maximum height is 0.97 seconds

`s=ut+(1)/(2)at^2\\\Rightarrow s=15* 1.52+(1)/(2)* -9.81* 1.52^2\\\Rightarrow s=11.47\ m`

So, the stone would travel 11.47 m up

So, total height stone would fall is 75+11.47 = 86.47 m

Total distance travelled by the stone would be 75+11.47+11.47 = 97.94 m

`s=ut+(1)/(2)at^2\\\Rightarrow 86.47=0t+(1)/(2)* 9.8* t^2\\\Rightarrow t=\sqrt{(86.47* 2)/(9.81)}\\\Rightarrow t=4.2\ s`

Time taken by the stone to travel 86.47 m to the water is is 4.2 seconds

The stone reaches the water after 4.2+1.52 = 5.72 seconds after throwing the stone

`v=u+at\\\Rightarrow v=0+9.81* 86.47 = 848.27\ m/s`

Speed just before hitting the water is 848.27 m/s