Question

An object thrown upward reaches the highest point in 5 sec. Find the velocity with which it was thrown? (Take g = 9.8 m/s2 )

Answer 1

as the time taken by the object to reach the highest point is 5 sec
let for example initial vel is u
so for every sec traveled by the object it's speed will decrease by 9.8 (as g=9.8m /s2)

also we know speed (final) at highest point is zero
therefore for 5 sec
u-5(9.8) = 0
u-49=0
u=49 m/s

Answer 2

Heya...

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ANSWER FOR YOUR QUESTION IS HERE:

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Time= t

Acceleration due to gravity = g

Distance = h

Initial velocity = u

Final velocity = v

t= 5

v= 0

g = -10 (As the object is thrown upwards)

By 1st equation

v=u+gt

0= u + (-10)*5

 = u + (-50)

u= 50 m/s

By 3rd equation

v^2 = u^2 + 2gh

v^2 - u^2 = 2gh

0^2 - (50)^2 = 2*(-10)*h

-2500= -20h

h= -2500/-20

 = 125m

Ans- Distance covered is 125 m

        Velocity with which object was thrown is 50 m/s

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Hope it helps

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