Question

A body is dropped freely under gravity from the top of a tower of height 78.4metre calculate (a) time to reach the ground.(b) velocity with which it will strike the ground .(acceleration=10m/s 2 )

Answer 1

time taken to reach the ground = 2.8×square root of 2 second
velocity with which it will strike the ground = 28×square root of 2 m/sec

Answer 2

Velocity with which it will strike the ground is 59.6 m/s

Solution:

The formula for calculating the time taken for a body to travel:

Time (t) taken by an object to fall freely from a distance (d) = t $= \frac{\sqrt{2 d}}{g}t= \sqrt{2} \times \frac{78.4}{9.8}=4 s$

Time is taken by the body to reach ground = 4s

Initial Velocity (u) = distance/time

$\mathrm{u}=\frac{78.4}{4}= 19.6 m/s$

Final velocity (v) = initial velocity (u) + acceleration (a) + time (t)

V = u + at

Given $a=10 \mathrm{m} / \mathrm{s}^{2}$

u= 19.6m/s

T = 4

V= u + at

V=19.6+$(10 \times 4)$= 59.6m/s

Velocity with which it will strike the ground is 59.6 m/s .