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Question

Physics Question on projectile motion

A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected at an angle of 6060^{\circ} with the vertical with the same initial speed. At highest points of their journey, the ratio of their potential energies will be

A

1:01

B

2:01

C

3:02

D

4:01

Answer

4:01

Explanation

Solution

For first ball At highest point, the kinetic energy is completely converted into its potential energy
mgh1=12mu2\therefore m g h_{1}=\frac{1}{2} m u^{2}
or h1=u22gh_{1}=\frac{u^{2}}{2 g}
For second ball
mgh2=mgu2sin2θ2gm g h_{2}=\frac{m g u^{2} \sin ^{2} \theta}{2 g}
=12mu2sin230=\frac{1}{2} m u^{2} \sin ^{2} 30^{\circ}
=12mu2(12)2=\frac{1}{2} m u^{2}\left(\frac{1}{2}\right)^{2}
or h2=u28gh_{2}=\frac{u^{2}}{8 g}
h1h2=u22g×8gu2\therefore \frac{h_{1}}{h_{2}}=\frac{u^{2}}{2 g} \times \frac{8 g}{u^{2}}
h1h2=41\Rightarrow \frac{h_{1}}{h_{2}}=\frac{4}{1}