Question

By what vertical velocity a piece of stone must be thrown,so as it may attain a maximum height of 1000m (g= 9.8m/s^2)​........
answer is 140m/s...can u plzz tell me the solution!!!!!!​

Answer 1

Answer:

Let u be the initial velocity and h be the maximum height attained by the stone.

v

1

2

=u

2

−2gh,

(10)

2

=u

2

−2×10×

2

h

100=u

2

−10h ....(i)

Again at height h,

v

2

2

=u

2

−2gh

(0)

2

=u

2

−2×10×h

u

2

=20h ... (ii)

So, from Eqs. (i) and (ii) we have

100=10h

h=10m

Answer 2

hey mate here is ur answer

At maximum height the velocity of stone will be 0m/s

In newton's third law of motion

v²-u²=2as

v=0(At max height)

s=1000m

0-u^2=2as

a=-g=-9.8(-ve because it's in upward direction)

-u²=-2(9.8)1000

u²=19600

u=140 m/s

hope it helps.

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