Question

A body is released from a height of 19.6 m. Find the velocity by which it strikes the ground
and also find time taken to reach the ground.​

Answer 1

Given

h=19.6m

u=0

g=9.8m/s²

To find

i.final velocity(v)

ii.time

Solution--

i.v²-u²=2gh

v²-0=2×9.8×19.6

v²=384.16

v=19.6m/s

ii.v=u+gt

19.6=0+9.8t

19.6=9.8t

t=19.6/9.8

t=2s

Thus,velocity by which it srikes the ground is 19.6m/s and time taken is 2 seconds.

Answer 2

Given :

• Height,h = 19.6m
• Acceleration, a = 9.8 m/s²
• Initial velocity,u = 0 m/s

To Find :

• The velocity will it strike the ground
• Time taken to reach the ground

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2as$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

${\underline{\sf{Answer}}}$

1) We have to Find the velocity by which the ball will strike the ground .

From the third equation of motion,we have:

$\sf\:v{}^{2}=u{}^{2}+2as$

Putting the values in the above formula,we get:

$\sf\:v{}^{2}={0}^{2}+2\times9.8\times19.6$

$\sf\:v{}^{2}=0+2\times192.08$

$\sf\:v{}^{2}=0+384.16$

$\sf\:v{}^{2}=384.16$

$\sf\:v=\sqrt{384.16}$

$\sf\:v=19.6ms{}^{-1}(approx)$

Thus, The velocity will it strike the ground is 19.6 m/s.

2)We have to find time taken to reach the ground.

From the equation of motion.

$\sf\:s=ut+\frac{1}{2}at{}^{2}$

Now put the given values

$\sf19.6=0\times\:t+\frac{1}{2}\times9.8\times\:t{}^{2}$

$\sf\:t^2=\dfrac{9.8\times2}{9.8}$

$\sf\:t^2=\dfrac{39.2}{9.8}$

$\sf\:t^2=4$

$\sf\:t=\sqrt{4}$

$\sf\:t=2\:sec$

Thus ,time taken to reach the ground is 2 sec.