Question

a child drop a stone from a cliff 49m height one second later the child throws a second stone after numerical problem for class 9

Answer 1

Answer:

Solution

Let time taken by thrown stone = t

Time taken by dropped stone = t+1

s =

1

2

g

(

t

+

1

)

2

substitute values of g= 9.81 and s= 49m

t =

√

10

-1

s =

1

2

g

(

t

+

1

)

2

= ut +

1

2

g

t

2

s = gt +

g

2

= ut

u = g(1 +

1

2

t)

= 12.08

m

s

e

c

Answer 2

Answer:

If time for first stone ist, time for second stone is t−1

s=49m is the same for both stones, u=0 for first stone

v 2 =u2 +2as

v 2 =0+2(9.8)(49)

v=31m/s

v=u+at

31=0+9.8t

t=3.16s

t ′ =2.16s

s=ut+0.5at 2

49=u′(2.16)+0.5(9.8)(2.16)2

u'=12.1m/s