A man stands on a cliff that is 10 m above the sea. He throws a stone vertically upwards with a velocity of [tex]5 \, \text{ms}^{-1}[/tex]. The stone eventually lands in the sea. If air resistance is negligible, what is the time taken for the stone to reach the sea after leaving the man's hand?
Answer (2)
Do you remember the general equation for the distance covered by a moving object ? There are not many perfect opportunities to use it in all its glory, but I think this is one of them.
Position = (starting distance) + (starting speed) (time) + (1/2) (acceleration) (time)²
H = starting position + (starting speed x t) + 1/2 A t²
Here's how we can use it, with some careful definitions:
-- Let's say the surface of the sea is zero height. Then 'H' ... the position at the end ... is zero, when it plunks into the water, and the starting, original position of the stone is +10 on the cliff in the man's hand.
-- Starting speed is +5 ... 5 m/s upward, when he tosses it.
-- Acceleration is 9.8 m/s² downward ... the acceleration of gravity.
I think this is going to work out just beautifully !
0 = (5) + 5t - 1/2 (9.8) t²
**-4.9 t² + 5t + 5 = 0 ** That's the whole thing right there. Look how gorgeous that is !
Solve it for 't' with the quadratic equation, A = -4.9 B = 5 C = 5
When you solve a quadratic with the formula, you always get two roots. If it's a real-world situation, one of them might not make sense. That's the result in this case.
The two roots are
t = - 0.622 second and t = + 1.642 second
The first one isn't useful, because it means 0.622 second before the man tossed the stone up.
So our answer is: We hear the 'plunk' 1.642 second after the upward toss.
The time taken for the stone to reach the sea after being thrown upwards is approximately 2.02 seconds. This is calculated using the kinematic equation for uniformly accelerated motion. The quadratic formula is used to solve for time, resulting in one valid positive solution.
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