Question

a ball is thrown vertically from the ground. It crosses a point at the height of 25 m twice at an interval of 4 sec.The ball was thrown with the velocity of

Answer 1

30m/s

2sec to reach the highest point where velocity will be zero(0) from the point 25 m above the ground .

V=u-gt

V=20m/s

Again, v^2-u^2=-2gs

V=20,u=?,s=25

u=30m/s

Answer 2

Hey friend here is ur solution..
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Given the ball is crosses Same point in 4 sec

Hence in 2 sec it reaches highest point from given point (at height=25m)

hence velocity at given point be v1=gt. ( v= v1-gt where v= 0)

velocity at height 25( v1)=9.8*2=19.6 m/s

now
Using v = u +at
let v1 =velocity at height 25=( v1)

and initial velocity=u

s=25.

v²=u²-2*g*h. (while ascending)

(19.6)²=u²-2*9.8*25

or u=29.56m/s

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Hope this will help you..