Question

A rock is thrown upward with a velocity of 17 meters per second from the top of a 31 meter high cliff and it missies the cliff on the way back down when will the rock ve 8 meters from the ground level

Answer 1

Answer:

$4.5\ \text{s}$

Step-by-step explanation:

t = Time taken

u = Initial velocity =17 m/s

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-17^2}{2\times -9.81}\\\Rightarrow s=14.73\ \text{m}$

Total height the rock is to fall when it is 8 meters from the ground is $14.73+(31-8)=37.73\ \text{m}$

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-17}{-9.81}\\\Rightarrow t=1.73\ \text{s}$

Time taken to reach the maximum height is 1.73 s.

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow 37.73=0+\dfrac{1}{2}\times 9.81t^2\\\Rightarrow t=\sqrt{\dfrac{37.76\times 2}{9.81}}\\\Rightarrow t=2.77\ \text{s}$

Time taken from the maximum height to the point which is 8 m from the ground is 2.77 s.

So, time taken from the moment of the throw to the point 8 m from the ground is $1.73+2.77=4.5\ \text{s}$.