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?

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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,...

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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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