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

Relax

Respuesta :

Answer:

h = 90.10 m

Explanation:

Given that,

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

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

The path followed by the projectile is given by :

[tex]h(t)=-4.9t^2+10t+85[/tex] ....(1)

For maximum height,

Put dh/dt = 0

So,

[tex]\dfrac{dh}{dt}=-9.8t+10=0\\\\t=\dfrac{10}{9.8}\\\\=1.02\ s[/tex]

Put the value of t in equation (1).

[tex]h(t)=-4.9(1.02)^2+10(1.02)+85\\\\=90.10\ m[/tex]

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