Question

A ball is dropped from a tower. In the last second of its motion it travels a distance of 15m. Find the height of the tower. (take g = 10 m/s)

Answer 1

Let S1 be the distance travelled at time (t).

And S2 to distance travelled at time (t-1).

it is provided that S1 - S2=15.And u=0➡S=ut+1/2gt^2➡S1-S2=15

➡(ut+1/2gt^2)-(ut+1/2g(t-1)^2)=15

a=10

0+1/2×10×(t^2-(t-1)^2)=15 5(t^2-t^2+2t-1)=15 10t-5=15

t=2 m.

➡S=ut+1/2gt^2 S=0+1/2×10×4 S=10×2=20 m