A ball is released from the top of a tower of height 50 m calculate its final velocity

Answer :

Given, u = 0 m/s

\( h = s = 19.6 m \)

g = 9.8 m/s2 (falling down)

We know that, \( v^2 – u^2 = 2 g s \)

\( \Rightarrow v^2 – (0)^2 = 2 x 9.8 x 19.6 \)

\(\Rightarrow \) v = 19.6 m/s

The final velocity just before touching the ground is 19.6 m/s.

Last updated at May 30, 2019 by

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NCERT Question 14 A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground. Since the stone is released from top, Initial Velocity = u = 0 m/s Distance travelled by the stone = Height of the tower = 19.6 m And, Acceleration = a = g = 9.8 m/s2 (Acceleration is positive because stone falls downwards) Finding final velocity (v) We know u, s and a, Finding v using 3rd equation of motion v2 − u2 = 2as v2 − (0) 2 = 2 × 9.8 × 19.6 v2 = 19.6 × 19.6 v2 = 19.62 v = 19.6 m/s Final velocity of the ball is 19.6 m/s

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Last updated at May 30, 2019 by Teachoo

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Answer

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Text Solution

Solution : Here, Initial velocity u=0 final velocity v=? (to be calculated ) Acceleration due to gravity `g=9.8 m//s^(2)` (Stone comes down ) And Height h=19.6 m Now ,we know that for a freely falling body : `v^(2)=u^(2)+2gh ` So , ` v^(2)=(0)^(2)+2xx9.8xx19.6` `v^(2)=19.6xx19.6` `v^(2)=(19.6)^(2)` v=19.6 m/s Thus the velocity of stone just before hitting the ground will be 19.6 metres per second .

Text Solution

Solution : According to the equation of motion under gravity. `v^(2)-u^(2)=2gs` Where, u=Initial velocity of the stone=o v=Final velocity of the stone s=Height of the stone =19.6m g-Acceleration due to gravity `=9.8ms^(-2)` Therefore `v^(-2)-o^(2)=2xx9.8xx19.6` `V^(2)=2xx9.8xx19.6=(19.6)^(2)` `V=19.6ms^(-1)` Hence, the velocity of the stone just before touching the ground is `19.6ms^(-1)`.