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puteri [66]
3 years ago
13

You toss a rock of mass mm vertically upward. Air resistance can be neglected. The rock reaches a maximum height hh above your h

and. What is the speed of the rock when it is at height h/4h/4?
Physics
2 answers:
Andrew [12]3 years ago
8 0

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}}

Substituteg=9.8m/s^2

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

skad [1K]3 years ago
3 0

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.

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Answer:

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