Question

Three particles are projected vertically upward from a point on the surface of the earth with velocities $\sqrt{\left(2gR/3\right)},\sqrt{\left(gR\right)},\sqrt{\left(4gR/3\right)}$ respectively where R is the radius of the earth and g is the acceleration due to gravity on the surface of the earth. The maximum heights attained are respectively $h_1{,h}_2,h_3$

• A. $h_1:h_2=3:4$
• B. $h_2{,h}_3=3:4$
• C. $h_1{,h}_3=1:4$
• D. $h_2=R$

Answer 1

The correct option is A $h_1:h_2=3:4$

Solution:-

At maximum height, $KE=0$ so,

$K{E}_i=\frac{1}{2}m{v}^2,$ $U_i=\frac{-GMm}{R}$

$K{E}_f=0$, $U_f=\frac{-GMm}{g}$

As grating is a conservative

force,

$K{E}_i+U_i=K{E}_f+U_f$

$\Rightarrow \frac{1}{2}m{v}^2-\frac{Gmm}{R}=\frac{-GMm}{h}$

$\Rightarrow h=\frac{-GMm}{\frac{1}{2}m{v}^2-\frac{GMm}{R}}=\frac{GM}{\frac{GM}{R}-\frac{V^2}{2}}$

So, $h_1=\frac{GM}{\frac{GM}{R}-\frac{1}{2}(\frac{2GM}{3R}{)}^{2/2}}=\frac{3}{2}R$

$h_2=\frac{GM}{\frac{GM}{R}-\frac{1}{2}(\frac{GM}{R}{)}^{2/2}}=2R$

$h_3=\frac{GM}{\frac{GM}{R}-\frac{1}{2}(\frac{4GM}{3R}{)}^{2/2}}-3R$

$h_1:h_2=3:4,h_2:h_3=2:3$ and $h_1:h_3=1:2$