Question

A boy throws a stone vertically upwards from a bridge is 20m above the surface is water.the stone strikes the water surface 8sec after its release.caluculate the speed by which the stone was thrown up?
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Answer 1

Answer:

Given:

The was thrown vertically upwards from a bridge which is 20 metres above water. The stone hits water after 8 sec.

To find:

Initial velocity of water ?

Calculation:

Even though the stone was thrown upwards, the displacement of the stone upon reaching the water surface will be 20 metres.

Applying EQUATIONS OF KINEMATICS:

h = ut + \dfrac{1}{2} a {t}^{2}h=ut+21at2

Here direction of displacement & gravity is same. However, direction of initial velocity is opposite.

\implies - 20 = u(8) + \dfrac{1}{2} ( - 10) {8}^{2}⟹−20=u(8)+21(−10)82

\implies - 20 = u(8) - 320⟹−20=u(8)−320

\implies u(8) = 320 - 20⟹u(8)=320−20

\implies u(8) = 300⟹u(8)=300

\implies u = 37.5 \: m {s}^{ - 1}⟹u=37.5ms−1

So, initial velocity of throwing the ball was 37.5 m/s

Answer 2

s = -20 m
a = g = -10 m/s^2
t = 8 sec

s = ut + 1/2 at^2
-20 = 8u - 320
8u = 300
u = 300/8
u = +37.5 m/s [Positive sign represents that the stone was thrown upwards]


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