Question

A body a of mass 4 kg is dropped from a height of 100m and another body b is dropped from a height of 50 m of 2 kg at same time then find the time taken by both the bodied to reach the ground

Answer 1

Answer:

Explanation:

In this question there is no use of mass,

For body 1:

Height = 100m

initial velocity = 0m/s

acceleration due to gravity = 10m/s^2

Therefore, we use the formula,

s = ut + 1/2 gt^2

100 = 1/2 10 t2

Therefore, t = 10v2seconds

For body2:

Height  = 50m

u = 0m/s

g = 10m/s^2

Using formula we get,

time = v10seconds

Hope this helps you...

Answer 2

Answer:

The body B takes nearly 0.7th of time required by A to reach the ground.

Explanation:

Body A dropped from height, $h_{A}=100m$

Body B dropped from a height, $h_{B}=50m$

From the 2nd equation of motion:

          $S=ut+\frac{1}{2}gt^{2}$

Initially body was at rest, u=0

Now      $h_{A}=\frac{1}{2}gt^{2}_{A}$    and  $h_{B}=\frac{1}{2}gt^{2}_{B}$

By dividing them we get,   $\frac{h_{A}}{h_{B}}=\frac{t^{2}_{A}}{t^{2}_{B}}$

                     $\frac{t_{A}}{t_{B}}=\sqrt{\frac{h_{A}}{h_{B}}}$

Put the values of heights in above equation:

                   $\frac{t_{A}}{t_{B}}=\sqrt{\frac{100}{50}}$

                    $t_{A}=\sqrt{2}t_{B}$

∴                  $t_{B}=0.7t_{A}$

Therefore  the time taken by body B is 0.7th of the time taken by body A.