A stone is dropped from the upper observation deck (the space deck) of the CN Tower, 450m above the ground.
(a) How long does it take the stone to hit the ground?
(b) If the stone is thrown downward with a speed of 5m/s, how long does it take to reach the ground?
To solve these questions, we can use the equations of motion under constant acceleration. The acceleration due to gravity near the Earth's surface is approximately 9.8 m/s², directed downwards. We will assume there is no air resistance.
(a) To find the time it takes for the stone to hit the ground when dropped from rest, we can use the equation:
h = (1/2)gt²
where h is the initial height (450 m), g is the acceleration due to gravity (9.8 m/s²), and t is the time it takes for the stone to hit the ground.
Rearranging the equation, we get:
t² = (2h) / g
Substituting the given values:
t² = (2 * 450 m) / (9.8 m/s²)
t² = 91.84
Taking the square root of both sides, we find:
t ≈ 9.58 s
Therefore, it takes approximately 9.58 seconds for the stone to hit the ground when dropped from the upper observation deck.
(b) If the stone is thrown downward with a speed of 5 m/s, we can use a similar approach. In this case, the initial velocity of the stone is -5 m/s (negative sign indicating it is directed downward).
Using the equation:
h = vit + (1/2)gt²
where h is the initial height (450 m), vi is the initial velocity (-5 m/s), g is the acceleration due to gravity (9.8 m/s²), and t is the time it takes for the stone to hit the ground.
Since the stone is thrown downward, h is positive and vi is negative. Rearranging the equation, we get:
t² + (2vi/g)t - 2h / g = 0
Plugging in the given values:
t² + (2 * -5 m/s / 9.8 m/s²)t - (2 * 450 m / 9.8 m/s²) = 0
Solving this quadratic equation will give us the time t.