Question

A ball is dropped from a height if take 0.2 seconds to cross the last 6 metres before hitting the ground then the height from which it was dropped{g=10m/s²}​

Answer 1

Given:-

• A ball dropped from height takes 0.2 seconds to cross the last six metre before it hits the ground.
• The value of acclⁿ due to gravity is 10m/s².

To Find:-

• The height from which the ball was dropped.

Formula Used:-

We will use second and third equⁿ of motion.

$\large\purple{\underline{\boxed{\pink{\tt{\dag s=ut+\dfrac{1}{2}at^2}}}}}$

$\large\purple{\underline{\boxed{\pink{\tt{\dag \:\:2as\:\:=\:\:v^2\:\:-\:\:u^2\:\:}}}}}$

Calculation:-

Let us take that the ball was dropped from a height h .

$\underline{\tt{\purple{\leadsto From\:B\:to\:C:}}}$

• Initial Velocity = ?
• Acclⁿ due to gravity = 10m/s².
• Distance travelled = 6 m.

So,on using 2nd equⁿ of motion we , have :

$\tt:\implies s=ut+\dfrac{1}{2}at^2$

$\tt:\implies 6m = u\times0.2s+\dfrac{1}{\cancel{2}}\times\cancel{10}\times(0.2)^2$

$\tt:\implies 6m = 0.2u + 5\times 0.04$

$\tt:\implies 0.2u = 6 - 0.2$

$\tt:\implies 0.2u=5.8$

$\tt:\implies u =\dfrac{\cancel{5.8}}{\cancel{2.9}}$

$\underline{\boxed{\red{\tt{\longmapsto u \:\:\:\:=\:\:\:\:29ms^{-1}}}}}$

_________________________________________

Again using 3rd equⁿ of motion , we have ,

• Initial Velocity = 0m/s.
• Final velocity = 29m/s.
• acclⁿ due to gravity = 10m/s².

$\tt:\implies 2as=v^2-u^2$

$\tt:\implies 2\times10m/s^2\times s=(29m/s)^2-(0m/s)^2$

$\tt:\implies 20s = 841m$

$\tt:\implies s=\dfrac{841m}{20}$

$\underline{\boxed{\red{\tt{\longmapsto s \:\:\:\:=\:\:\:\:42.05m}}}}$

$<font color = baby >$

Now it earlier mentioned , it covered 6m in last 0.2s .

Hence total distance covered = 42.05m+6m= 48.05m.

$\underline{\tt{\pink{\leadsto Hence\:the\:height\:from\:it\:was\: dropped\:is\:48.05m.}}}$