Question #01e2a

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Question #01e2a

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Answer 1 · 2018-03-01T13:44:24

We can use $v = u + a t$ $v=u+at$ to calculate speed of the ball after $t s$ $ts$ ,

here, $u = 4 m s^{- 1}, a = g, t = 2$ $u=4ms^-1,a=g,t=2$

So, $v = 4 + 2 \cdot 9.8 = 23.6 m s^{- 1}$ $v=4+2*9.8=23.6ms^-1$

And,if its speed is $V$ $V$ while touching the ground,then we can use, $v^{2} = u^{2} + 2 g h$ $v^2 =u^2 +2gh$

here, $h = 16 m$ $h=16m$

So, $V = \sqrt{16 + 2 \cdot 16 \cdot 9.8} = 18.157 m s^{- 1}$ $V =sqrt(16+2*16*9.8)=18.157ms^-1$

Oops! The velocity value while touching the ground is less than the velocity found at $2 s$ $2s$ ,that means the ball has reached the ground before that.

Let's calculate time required to reach the ground,using $V = u + g t$ $V=u+g t$

So, $t = \frac{18.157 - 4}{9.8} = 1.45 s$ $t=(18.157-4)/9.8=1.45 s$

So,now if we consider that after the impact the ball went upwards,and also consider that coefficient of restitution was $1$ $1$ ,then that ball will go up with the same velocity.

So,for $(2 - 1.45) = 0.55 s$ $(2-1.45)=0.55s$ while going up,if it achieves a velocity of $v$ $v$ ,then we cans say,

$v = u - g t$ $v=u-g t$

here, $u = 18.157 m s^{- 1}$ $u=18.157ms^-1$

So, $v = 12.767 m s^{- 1}$ $v=12.767ms^-1$ (upwards)(this is the velocity at the end of $2 s$ $2s$ )

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