Falling and bouncing ball problem

Question

I have this interesting physics problem I can't solve. Could you please explain it in simple terms?

A small ball thrown vertically down from height of 4,9 m with initial speed of 9,8 m/s hit a horizontal motionless platform and bounced back to the same height of 4,9 m. 

1. Find the speed the ball reached the platform and the speed the ball bounced from the platform. Provide the given answers with the accuracy of 1 cm/s. 

2. What part of ball's mechanical energy (in percent) was lost during the collision? Provide the given answer by rounding to the tenth part of a second.

Don't care about air resistance and duration of the first collision.

Answer 1

I'll use the acceleration due to gravity (g) as 9.8 m/s². I'll define 4 times, Time 1 is the initial condition, Time 2 is just instantaneously before the bounce and Time 3 is just instantaneously after the bounce. Time 4 is when the ball reaches the 4.9 m mark at the end.

The ball, initially, had energy consisting of both potential (PE = mgh) and kinetic (KE = 1/2mv²) so we can so the Total Mechanical Energy (TME) is the sum of the two:

(I'll let TME₁ by the Total Mechanical Energy at time 1 or the initial condition)

TME₁ = PE₁ + KE₁

TME₁ = mgh + 1/2mv²

TME₁ = m(9.8)(4.9) + 1/2m(9.8²)

TME₁ = 48.02m + 48.02m = 96.04m

Instantaneously prior to the ball hitting the platform, the ball had a TME we'll define as TME₂ as follows:

TME₂ = PE₂ + KE₂

TME₂ = mgh + 1/2mv²

TME₂ = m(9.8)(0) + (1/2)(m)(v²)

TME₂ = (1/2)(m)(v²)

Conservation of energy tells us that E₁ = E₂ and since we are only dealing with mechanical energy TME₁ = TME₂

TME₁ = 96.04m

TME₂ = (1/2)(m)(v²)

TME₁ = TME₂ so 96.04m = (1/2)(m)(v²)

Divide both sides by 1/2m to get:

192.08 = v²

v = √192.08 = 13.86 m/s or 1386 cm/s

For now, we will skip time 3 which is just instantaneously after the ball bounces and move to time 4 which is when the ball reaches the 4.9 meter mark.

TME₄ = PE₄ + KE₄

TME₄ = mgh + 1/2mv²

TME₄ = m(9.8)(4.9) + 1/2m(0)²

TME₄ = 48.02m

The amount of energy lost during the bounce is TME₄ - TME₂ = 48.02m - 96.04m = -48.02m (the negative is the same as saying energy LOST)

The percentage of energy lost is (TME₄ - TME₂)/TME₂ = 48.02m/96.04m = 50% (I don't understand "Provide the given answer by rounding to the tenth part of a second." because percentage does not have units; seconds is not involved but you could answer 50.0%)

Now, let's look at the TME₃. During the bounce, energy was lost. Instantaneously after bouncing, the TME was the same as at time 4 so TME₃ = TME₄ = 48.02m

But TME₃ = PE₃ + KE₃

TME₃ = mgh + 1/2mv²

TME₃ = m(9.8)(0) + 1/2mv²

TME₃ = 1/2mv²

48.02m = 1/2mv²

v² = 96.04

v = √96.04 = 9.80 m/s or 980 cm/s (The velocity at time 3, instantaneously after the bounce)