Question

A ball is released from the top of a tower of height 'h' meters. It takes 'T' seconds to reach the ground.What is the position of the ball in T/3 seconds?

a) h/9 metres from ground.
b)7h/9 m from ground.
c) 8h/9 m from ground.
d) 17h/18 m from ground.

Answer 1

ok, so using the position function, you see that
\[\Delta y = .5 g t^2 \] and
\[\Delta y = .5g (t/3)^2 = (.5 g t^2) / 9\] when t =t/3

so essentially, you have, h-h/9= 8h/9

Answer 2

any short method very high level method.. @Inspired

Answer 3

@UnkleRhaukus @phi @Shadowys plz help

Answer 4

that method is quite quick and high level enough. :)

Answer 5

but i am in 9 class can you tell me step by step

Answer 6

the motion formula is \(S=ut + \frac{1}{2} at^2\), as u=0, and a=g, it simplifies to \(h=\frac{1}{2} aT^2\)
when u=0, t=T/3,
sub it in to get \(S= \frac{1}{2} a\frac{T^2}{9}=\frac{\frac{1}{2} aT^2}{9}=h/9\)
subtracting the remaining height givens 8h/9