# A ball is thrown downwards from the top of a 40.0 m tower. It takes 2 sec to hit the floor. What is its initial and final speed?

We can use equations of motion to solve this problem. Let us first solve for the initial speed of the ball, by using the following equation:

s = ut + a/2 at^2

here, s is the distance traveled (= height of tower = 40 m), u is the initial speed,...

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We can use equations of motion to solve this problem. Let us first solve for the initial speed of the ball, by using the following equation:

s = ut + a/2 at^2

here, s is the distance traveled (= height of tower = 40 m), u is the initial speed, t is the time taken (= 2 s) and a is the acceleration of the ball and is equal to acceleration due to gravity or g.

Substituting these values in the equation, we get:

40 = 2 x u + 1/2 x 9.81 x 2^2

solving this, we get: u = 10.19 m/s

Using  v = u + at

v = 10.19 + 9.81 x 2 = 29.79 m/s

Thus, the ball was thrown with an initial speed of 10.19 m/s and reached the ground surface at 29.79 m/s.

Hope this helps.

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