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?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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,...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

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. 

Approved by eNotes Editorial Team