Question

A ball is dropped from the top of a building 19.6 m high.find the time taken by the ball to reach the ground.

Answer 1

The ball at rest, before being dropped, has a potential energy due to gravity that depends on its height. Its kinetic energy is zero, because it is at rest.Just before it hits the ground, its potential energy is zero (its height is zero), but now it has kinetic energy due to its motion.Since energy is conserved, the potential energy from step 1 must equal the kinetic energy from step 2.You should know (or be able to Google) the expressions for potential energy in a gravitational field, and kinetic energy due to motion.Set those two expressions to be equal. They'll both include a factor for the mass of the ball, but you should find that the mass cancels out- it appears on both sides of the equation.Solve that equation for v (velocity)- it should be some expression involving g (the acceleration due to Earth’s gravity- also easily found online, but commit it to memory! Make sure you express it in m/(s^2) ) and h (the height at which the ball was released in meters).Profit from working through the problem yourself, and hopefully learning something along the way!

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Answer 2

S = 19.6

Initial velocity (u) is 0

acceleration due to gravity = 10N

Now we use the formula of equations of motion

s = ut + 1/2 at^2

So 19.6 = 0 + 1/2 × 10t^2
5t^2 = 19.6
t^2 = 196/50 = 3.92

t = √3.92

t = 1.97 seconds

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