A stone is dropped from the upper observation deck (the Space Deck) of the CN Tower, 450m above the ground.

Question

a) Find position function s(t) of the stone above the ground level at time t. Assume g=9.8 m/s².

b) How long does it take the stone to reach the ground?

c) What is the impact velocity of the stone when it hits the ground?

d) If the stone is thrown downward with a speed of 5m/s. How long does it take to reach the g

Answer 1

Since the acceleration of gravity can be considered a constant for this situation the below formulas can be derived:

1) acceleration = g = constant = -9.81 m/s² dv/dt

2) velocity =v₀ -gt = ds/dt

3) Position s=s₀+v₀t -1/2gt²

a) Therefore in this problem s₀= 450 m and v₀ = 0, s(t) = 450 -1/2gt²

b) The time to reach the ground is found by solving the position equation for s=0 yielding 9.58 sec

c) The impact velocity is found by solving 2) for t= 9.58 sec yielding -94 m/s ( neg means down)

d) If the initial velocity is -5 m/s, find the time to reach the ground by solving 3) with v₀ = -5 yielding 9.05 sec