Question

6. A student drops a ball from the top of the tower of height of 19.6m. The
velocity with which the ball hits the ground is :
A 9 8m/sec
B 12/8m/sec
C. 19.6m/sec
D. 39.2m/sec​

Answer 1

Answer:

19.6m/s

Explanation:

Given :-

Tower height (distance traveled by the stone) , s = 19.6m

Initial velocity, u = 0m/s

Accelaration,a = g = 9.8m/s^2

To Find :-

Velocity with which the ball hits the ground (Final velocity), v = ?

Formula to be use :-

Third equation of motion :

$\sf{}v^2-u^2=2as$

Solution :-

$\implies \sf{}v^2-0=2(9.8)(19.6)$

$\implies \sf{}v^2=(19.6)(19.6)$

$\implies \sf{}v^2=(19.6)^2$

Take square root on both the sides

$\implies \sf{}v=19.6m/s$

Therefore, (option c) The velocity with which the ball hits the ground is 19.6m/s

Equation of motion :-

Relation among velocity,time,distance,acceleration is called equation of motion.

There are three equation of motion :-

$\sf{} First\ equation\ of\ motion :v =u+at$

$\sf{} Second\ equation\ of\ motion :s =ut+\dfrac{1}{2}at^2$

$\sf{} Third\ equation\ of\ motion : \sf{}v^2-u^2=2as$

Answer 2

Answer:

19.6 m/sec

Explanation:

v^2 - u^2 = 2gh

u=0

v^2 = 2× 9.8 × 19.6

=> v×v = 19.6× 19.6

=> v= 19.6 m/s