(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A ball of mass 0.5 kg rolls off a table 4 metres above the ground with a speed of 4 ms^-1.

a) what is the kinetic energy of the ball when it strikes the ground.

b) what is the kinetic energy of the ball just before it strikes the ground

c) what is the speed of the ball as it strikes the ground

2. Relevant equations

Ek=1/2mv^2

V^2=u^2+2as

Ep=mgh

3. The attempt at a solution

okay so to get a) I did Ep=mgh so when the ball strikes the ground, all Ep should be turned into Ek right?

So 0.5*9.8*4=19.6 J

for b) I know the horizontal final velocity component is 4 ms^-1. The vertical final velocity should be found using v^2=u^2+2as.

v^2=2*-9.8*-4

v=8.85 ms^-1

using pythagoras throem I got the final v to be 9.7159 ms^-1.

Ek=9.715^2*0.5*0.5=23.6 J

What I'm not sure about is for answer b) I got 23.6 and for a) I got 19.6. Shouldn't a) be a larger value than b) because with a), the ball has striked the ground. I don't know where I have gone wrong

please help

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# Homework Help: Kinetic energy of a ball when falling (projectile motion)

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