Question

from what height should a body of mass 40kg fall to have same kinetic energy as a body of mass 1.96kg travel at 12m/s

Answer 1

First take out the velocity of particle
0.5×40×v^2 = 0.5 ×1.96×144
V^2 (approx) = 7
We know v=2gh
H=v/2g
H=7/20
H=0.35m

Answer 2

Answer :

h = 0.36 m

Explanation :

At some point of time in the free fall,their Kinetic Energies are supposed to be equal

So,

$\sf{K = k}$

To finD :

Height from which the body should be dropped so that their Kinetic Energies are equal

Given :

Object I :

• Mass,m = 40 Kg

• Velocity,v = ?

Object II :

• Mass,M = 1.96 Kg

• Velocity,V = 12 m/s

Mathematically,Kinetic Energy is given by :

$\boxed{ \boxed{ \rm{K = \frac{1}{2} M {v}^{2} }}}$

Putting the values,we get :

$\longrightarrow \: \sf{ \cancel{ \dfrac{1}{2}} \times 40 \times {v}^{2} = \cancel{ \dfrac{1}{2}} \times 1.96 \times {12}^{2} } \\ \\ \longrightarrow \: \sf{ {v}^{2} = \dfrac{1.96 \times 144}{40}} \\ \\ \longrightarrow \: \underline{ \boxed{ \sf{ {v}^{2} = 7.05 \: {ms}^{ - 1} }}}$

From the relation,

$\boxed{ \boxed{ \rm{ {v}^{2} - {u}^{2} = 2gh }}}$

Assuming that the initial velocity would be zero and g = 10 m/s²

$\longmapsto \: \sf{7.05 - {0}^{2} = 20h } \\ \\ \longmapsto \: \sf{h = \dfrac{7.05}{20}} \\ \\ \longmapsto \: \underline{ \boxed{ \sf{h = 0.35 \: m \: }}}$