Question : A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds.
Doubt by Jaskirat
Solution :
When the ball is released from the toper of a tower then
s1=h
t1=T
Using 2nd Equation of Motion
s=ut+½at²
s1=ut1½g(t1)²
h=(0)T+½g(T)²
h=½gT² — (1)
s1=ut1½g(t1)²
h=(0)T+½g(T)²
h=½gT² — (1)
s2=?
t2=T/3
Using 2nd Equation of Motion
s=ut+½at²
s2=ut2½g(t2)²
s2=(0)T+½g(T/3)²
s2=½g(T²/9) — (2)
s2=ut2½g(t2)²
s2=(0)T+½g(T/3)²
s2=½g(T²/9) — (2)
Dividing Equation (2) by (1)
s2/h =[½g(T²/9)]/[½gT²]
s2/h=1/9
s2=h/9
s2/h=1/9
s2=h/9
The position of the ball from the point of release in T/3 seconds is h/9 metre.
Also, position of the ball from the bottom in T/3 will be
=h-h/9
=8h/9 metre.
Similar Question :
A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?
a) h/9 metres from the ground.
b) 7h/9 from the ground.
c) 8h/9 metres from the ground.
d) 17h/18 metres from the ground.