Question

A stone released from a certain height ‘h reaches the ground in time 4 seconds.
When will it be at / h below the point of release? (g = 10 m s').
(a) 2 seconds
(b) 1 second
(c) 2V2 seconds
(d) V3 second
11​

Answer 1

Answer:

Explanation:

4=10×h

4/10=h

2/5=h

Let's approx and make it 2V2

Answer 2

Answer:

The stone will be at 2.83s below the point of release.

Explanation:

A stone is released from a certain height H and reaches the ground in 4 seconds. When will it be at H/2 below the point of release?

Let T denote the time (in s) taken by the stone to be at H/2 below the point of release

$H=0+(\frac{1}{2} )*g*4^{2} =8g\\H/2=0+(\frac{1}{2} *g*T^{2} =\frac{1}{2} *g*T^{2}$

Now,

$T^{2} =\frac{H}{g} = 8/T=\sqrt{8} = 2.83s$

#SPJ3