A mountain climber stands at the top of a 50 meter high cliff that overhangs a c

Question

A mountain climber stands at the top of a 50 meter high cliff that overhangs a calm pool of water. He throws two stones vertically downward 1.00 second apart and observes that they hit the water at exactly the same time. The first stone had an initial velocity of -2.00 m/s. How long after release of the first stone did the two stones hit the water?


Here's what you wrote before:

To find the time it took the first stone to hit:

d = Vi*t + (1/2)*a*t^2
-50 = -2*t + (1/2)*(-9.8)*t^2

using the quadratic equation, t = 3.00 seconds


My question is why would the change in x be -50 rather than positive 50? And wouldn't the acceleration be positive since it's going towards the earth and therefore speeding up?

Explanation / Answer

take +y direction upwards now the stone is moving downwards and the gravitational acceleration is also doenwards so both are negative and as the initial speed is also downwards so acceleration and velocity are in same direction so the stone will speed up downwards.

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