A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.

If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.

Solve for Vmax.

Answer / Explanation

It is worthy to note that the question is incomplete. There is a part of the question that gave us the vale of V₀.

So for proper understanding, the two parts of the question will be highlighted.

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.

a) What must the height of the building be for both balls to reach the ground at the same time if (i) V₀ is 6.0 m/s and (ii) V₀ is 9.5 m/s?

b) If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.

Solve for Vmax

Step Process

a) Where h = 1/2g [ (1/2g - V₀) ² ] / [ (g - V₀) ²]

Where V₀ = 6m/s,

We have,

h = 4.9 [ (4.9 - 6) ²] / [ (9.8 - 6) ²]

= 0.411 m

Where V₀ = 9.5m/s

We have,

h = 4.9 [ (4.9 - 9.5) ²] / [ (9.8 - 9.5) ²]

= 1152 m

b) From the expression above, we got to realise that h is a function of V₀, therefore, the denominator can not be zero.

Consequentially, as V₀ approaches 9.8m/s, h approaches infinity.

Therefore Vₙ = V₀max = 9.8 m/s