A basketball player throws his ball vertically upwards with an initial speed of v=40 m/s. Ignore air resistance. What is the speed of the ball at half of the maximum height?
Since we are neglecting air resistance the energy of the ball is conserved. We set the gravitational potential energy to be U=0 at h=0. Applying conservation of energy at h=0 and h=hmax , we get: U1 + K1=U2 +K2(1), at h=0 the potential energy is U1=0 since we did set it so and at the maximum height the speed is 0, therefore K2=0. So, (1) becomes mv2/2=mghmax (2).
Now applying conservation of energy at h=0 and h=hmax/2: mu2/2+mghmax/2=mv2/2, and using (2) we get, mu2/2 +mv2/4=mv2/2, which simplifies to: u2=v2/2, therefore the speed of the ball at h=hmax /2 is u=v/sqrt(2)=28.28 m/s
AM
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