Question

A ball is dropped from the top of a straight cliff. the ball reaches the ground 4second later. how high is the cliff? what is the final velocity of the ball when it just reaches the ground? (take g= 10 m s^-2​

Answer 1

The general formula for this is:

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→ Distance_2 = 1/2 * Acceleration * Time * Time + Initial Velocity * Time + Distance_1

→ Since you didn’t mention an initial velocity, let’s assume that’s Zero (0).

→ You asked for height above ground (Distance_1) and Distance_2 is Zero (0), height at ground.

→ Acceleration due to gravity at sea level (assuming this is at sea level) is -9.8 meters/second/second. We represent this by g.

→ Filling in the Zeros and using H for height above ground, our new formula is:

= 0 = 1/2 * g * t^2 + 0 * t + H

Simplifying and rewriting that, we get:

H = -1/2 * g * t^2

t = 5 seconds

g = -9.8 meters/second/second

→ Plug in the numbers for t and g and solve for H. Be careful with the negative signs!

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