A 80.0 g egg dropped from a window is caught by a student. If the student exerts a net force of −2.5 N over a period of 0.25 s to bring the egg to a stop, what is the egg’s initial speed?

Known :

m = 80 gr = 0.08 kg

F = 2.5 N

∆t = 0.25 s

Solution :

F . ∆t = ∆P

F . ∆t = m(v₂ - v₁)

(-2.5 N)(0.25 s) = (0.08 kg)(0 - v₁)

v₁ = 7.8125 m/s

Here's what I get  

Explanation:

A. Distance between A and B.

h = -½gt²

The stones go faster the farther they fall.

Stone A has already reached 5 m when B is released.

When B reaches 5 m, A has dropped further and is falling even faster.

The distance between the stones increases with time.

Figure 1 shows this effect in a graph of height vs. time.

B. Speed of Stone B

v² = 2gh =2 × ( -9.81 m·s⁻²) × (-5 m) = 98.1 m·s⁻²  

v = 9.9 m/s

The stone is travelling at 9.9 m/s when it reaches 5 m.

C. Velocity vs time

v = -gt

Both stones accelerate at the same rate.

When Stone B has reached 10 m at time t, Stone A is falling much faster.

Fig. 2 shows this in a graph of velocity vs time.