What is the speed u of the object at the height of (1/2)hmax? At what height h a

Question

What is the speed u of the object at the height of (1/2)hmax? At what height h above the ground does the projectile have a speed of 0.5v? Using conservation of energy, find the maximum height hmax to which the object will rise. A ball is launched with initial speed v from ground level up a frictionless slope (This means the ball slides up the slope without rolling). The slope makes an angle with the horizontal. Using conservation of energy, find the maximum vertical height hmax to which the ball will climb.

Explanation / Answer

According to the law conversation of energy

mgh = mu2/2

hmax= u2/2g

hmax/2 = u2/4gg

From 3 rd law motion

V2 - u2 = -2gh

0.25u2-u2 = - 2gh

h = 3u2/8g = 3hmax /4

The final velocity don't depend on angle inclined plane

Hence h = v2 /2g

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