1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A body is dropped from a certain height at time T=0. It undergoes freefall and reaches the ground at time T=t s. What is the expression for the height h from which the body started its free fall? Acceleration due to gravity is g.

A

h=gt2

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B

h=gt

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C

h=12gt2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

h=(12)gt2

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is C h=12gt2 The distance covered with initial velocity u and undergiong constant accleration a in time t is given by: s=ut+12at2 In this case the distance travelled is h, acceleration is g We know that h=ut+(12)gt2. Since the body is initially at rest, u=0. Therefore, h=(12)gt2

Suggest Corrections
3
Join BYJU'S Learning Program
Related Videos
Free Fall
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program