Question

a man is standing near the edge of a cliff 85 meters high. he throws a stone upward vertically with an intial velocity of 10 m/s. the stone clears the cliff edge on the way down and falls all the way to the ground. what is the maximum height of the stone above the ground

Answer 1

Answer:

h = 90.10 m

Explanation:

Given that,

A man is standing near the edge of a cliff 85 meters high, h₀ = 85 m

The initial speed of the stone, u = 10 m/s

The path followed by the projectile is given by :

$h(t)=-4.9t^2+10t+85$ ....(1)

For maximum height,

Put dh/dt = 0

So,

$\dfrac{dh}{dt}=-9.8t+10=0\\\\t=\dfrac{10}{9.8}\\\\=1.02\ s$

Put the value of t in equation (1).

$h(t)=-4.9(1.02)^2+10(1.02)+85\\\\=90.10\ m$

So, the maximum height of the stone is equal to 90.10 m.