Question

A particle, starting from rest, falls 70 metres in the last second of its motion. determine the total time taken by particle to fall, and the height from which it fell.

Answer 1

that's the way how to get height

Answer 2

Answer:

Total time taken by particle to fall is 6.5 seconds and height from which it fell is 211.25m.

Explanation:

Given is a particle which falls by 70m in its last second.

Let the total time taken by the particle to fall be T,

height fallen by the particle in T seconds

$H_{1}=\frac{1}{2}\times g\times T^{2} \\where g=9.8m/s^{2}$

height fallen by the particle in (T-1) seconds,

$H_{2} =\frac{1}{2}\times g\times (T-1)^{2}$

So the height fallen in the last second is,

$H_{1}-H_{2} = 70m =\frac{1}{2}\times g\times (T)^{2}-\frac{1}{2}\times g\times (T-1)^{2}$

On solving for T we get,

T= 6.5seconds

Hence, the total time take by the particle to fall is 6.5 seconds.

Total height covered is given by,

$H=uT+\frac{1}{2}\times g\times T^{2}\\ Since\ u=0,\\ H=\frac{1}{2}\times g\times T^{2}\\H=\frac{1}{2}\times 10\times (6.5)^{2}\\H=211.25m$

Hence the height from which it fell is 211.25m.