Question

A body dropped from top of a tower fall through 40m during the last two seconds of its fall. The height of the tower is (g=10m/s^{​​​​​​2}).

A) 60m

B) 45m

C) 80m

D) 50m

Answer 1

Let h = the height of the tower that needs to be determined.

Let t be the time of fall.

Then (t - 2) would be the time to reach the top of the 40 meter mark, and let d be the distance fallen till it reaches the 40 m mark.

Using kinematics, we can write: d = 1/2 g (t-2)^2

And H = 1/2 g t^2

H also = 40 + d

Then: H = 40 + 1/2 g (t-2)^2 = 1/2 g t^2

Expand: 40 +5 (t^2 - 4 t + 4) = 5 t^2

40 + 5 t^2 - 20 t + 20 = 5 t^2

Add like terms: - 20 t = - 60

t = 3 s

(t-2) = 1 second

In one second an object falls 5 m

Then H = 45 m