Question

A stone is dropped from top of the tower of height " h". After 1 second another stone is dropped from the balcony 20m below the top.
Both reach the bottom simultaneously. what is the value of h? taking g=10m/sec2

Answer 1

Given,

Initial velocity, u = 0 (In both the cases)

Height of tower = h

Height of balcony = h − 20

When stone is dropped from the tower, suppose the time taken is t, then for the second

case it will be t − 1.

Thus using equation of motion,

h = ut + = 1/2gt raise to 2=0+ 1/2gt raise to 2------------------(i)

and h − 20 =1/2g(t-1)raise to 2--------------------------------------(ii)

Solving (i) and (ii), we get

t = 2. 5 s

and h = 31. 25 m

Answer 2

Answer:

3.25 m

Explanation:

Let the time be t taken by first stone to reach ground from height h.

A/q

S=1/2 gt²

s=1/2x10t²

Let the time be t taken by first stone to reach ground from height h.

S=1/2 gt²

(h-20)=1/2x10(t-1)²

soving above 2 equations

5t²-20=5(t²-2t+1)

t=2.5 sec

now

h=1/2 x 10 x 2.5 x 2.5²=31.25 m

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