Question

You toss a rock of mass mm vertically upward. Air resistance can be neglected. The rock reaches a maximum height hh above your hand. What is the speed of the rock when it is at height h/4h/4?

Answer 1

Answer:

The speed of the rock when it is at height h/4 is $\dfrac{\sqrt{3gh} }{2}$.

Explanation:

At maximum height the final velocity of the rock is equal to 0. Let u is the initial velocity of the rock. Using the conservation of energy to find it as :

$u^2=2gh$.......(1)

We need to find the speed of the rock when it is at height h/4. Let v' is the speed. Using 3rd equation of motion as :

$v'^2=u^2+2as$

here a = -g and s = h/4

$v'^2=u^2-2g\times \dfrac{h}{4}$

Using equation (1) :

$v'^2=(2gh)-2g\times \dfrac{h}{4}\\\\v'^2=\dfrac{3gh}{4}\\\\v'=\dfrac{\sqrt{3gh} }{2}$

So, the speed of the rock when it is at height h/4 is $\dfrac{\sqrt{3gh} }{2}$. Hence, this is the required solution.

Answer 2

Answer:

$3.83\sqrt{h} m/s$

Explanation:

We are given that

Mass of rock=m

We have to find the speed of the rock when it is at height h/4

When the rock reaches at maximum height then speed,v=0

Let u be the initial velocity of rock=u

We know that

$v^2-u^2=-2gh$

$0-u^2=-2gh$

$-u^2=-2gh$

$u^2=2gh$

When h=h/4

$v'^2-u^2=-2g\times \frac{h}{4}=-\frac{gh}{2}$

$v'^2-2gh=-\frac{gh}{2}$

$v'^2=2gh-\frac{gh}{2}=\frac{4gh-gh}{2}=\frac{3gh}{2}$

$v'=\sqrt{\frac{3gh}{2}}$

Substitute$g=9.8m/s^2$

$v'=\sqrt{\frac{3\times 9.8h}{2}}=3.83\sqrt h$ m/s