At the moment when a shotputter releases a 5 00 kg shot, the shot is 3 00 m above the - Physics - Work Energy And Power

Question

At the moment when a shotputter releases a 5 00 kg shot, the
shot is 3 00 m above the ground and
travelling at 15 0 m/s It reaches a maximum height of 8 00 m
above the ground and then falls to the
ground If air resistance is negligible,
What was the kinetic energy of the shot as it left the
hand?

What was the kinetic energy of the shot at its maximum
height?
What was the kinetic energy of the shot just as it struck the
ground?

Vijay Yadav · Accepted Answer

Given u = 15m/s

Maximum height reached by shot, H = 5m

Mass of shot, m = 5kg

Height of shot from the ground, h = 3m

Assume shot makes an angle θ with the horizontal.

$H = \frac{u^{2} \sin^{2} θ}{2 g} = > 5 = \frac{15^{2} \sin^{2} θ}{2 * 10} = > \sin^{2} θ = \frac{4}{9} = > s i n θ = \frac{2}{3}$

(a) kinetic energy of the shot as it left the hand = 0.5mu²

=> KE = 0.5*5*15*15 = 562.5J

(b) At the maximum height shot has only horizontal velocity = ucosθ = 15cosθ

So,

$K E = 0.5 * 5 * (15 \cos θ)^{2} = > K E = 2.5 * (15 * \frac{\sqrt{5}}{3})^{2} = > K E = 2.5 * 225 * 5 / 9 = 312.5 J$

(c) The kinetic energy of the shot just as it struck the ground K' =?

Horizontal velocity remains same through out the motion.

So, vertical velocity at point of struck is given by

${v_{y}}^{2} = (u \sin θ)^{2} + 2 g h = > {v_{y}}^{2} = (10)^{2} + 2 * 10 * 3 = 160 = > v_{y} = 12.65 m / s S o, v' = \sqrt{{v_{y}}^{2} + (u \cos θ)^{2}} = > v' = 16.88 m / s$

K' = 0.5*5*285 = 712.5 J

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