Question

A stone is released from the top of a tower of height 19.6 m.
Calculate its final velocity just before touching the ground.
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Answer 1

Answer:

0

Explanation:

let mark A on the left side.the tower is 90o.so tan90= 1.96/A

1=1.96/A

=1.96

hope it will help you

Answer 2

for the stone,

height fallen, h = 19.6 m

initial velocity, u = 0 m/s

final velocity just before touching the ground = v

we know that

the third kinematical equation for linear motion is

${v}^{2} = {u}^{2} + 2gh \\ {v}^{2} = {(0)}^{2} + 2(10)(19.6) \\ {v}^{2} = 2(196) = 392 \\ v = \sqrt{392} = \sqrt{2 \times 2 \times 2 \times 7 \times 7} \\ v = \sqrt{ {7}^{2} \times {2}^{2} \times 2 } = 7 \times 2 \times \sqrt{2} \\ v = 14 \sqrt{2} \: m {s}^{ - 1}$

Hence, velocity of the stone just before touching the ground is 14√2 m/s.