Question

If a coin is tossed up with velocity 21 ms
In air,
which returns back't' sec. Find its final velocity it
returns back to the initial point
Option A
ms)
21 ms
Option :B
15 ms
)
Option :C
10.5 ms
Ans A
option ...reason​

Answer 1

Answer:

$u = 21 \\ v = 0 \\ a = - 10 \\ \\ {v}^{2} = {u}^{2} + 2as \\ {(0)}^{2} = {(21)}^{2} + 2( - 10)(s) \\ - 441 = - 20s \\ \frac{441}{20} = s \\ u = 0 \\ s = \frac{441}{20} \\ a = 10 \\ {v}^{2} = {u}^{2} + 2as \\ {v}^{2} = 2(10)( \frac{441}{20} ) \\ {v}^{2} = 20 \times \frac{441}{20} \\ {v}^{2} = 441 \\ v = 21$

in upward motion initial velocity is 21 and final is zero, because when the coin reaches the height the final velocity becomes zero and acceleration is -10as the coin as gravity is acting in opposite direction. And in downward direction the initial velocity is zero and acceleration is 10.

hope it works