Question

The acceleration due to gravity on the planet $A$ is $9$ times the acceleration due to gravity on planet $B$ . A man jumps to a height of $2 \, m$ on the surface of $A$ . What is the height of jump by the same on the planet $B$ ?

• A. $6m$
• B. $\frac{2}{3} \, m$
• C. $\frac{2}{9} \, m$
• D. $18 \, m$

Answer 1

Answer: (D)

It is given that, acceleration due to gravity on plane $A$ is $9$ times the acceleration due to gravity on the planet $B$ , i.e.,
$g_{A}=9g_{B}$ .... (i)
From the 3rd equation of motion, we have
$v^{2}=2gh$
At planet $A$ ,
$h_{A}=\frac{v^{2}}{2 g_{A}}$ .... (ii)
At planet $B$ ,
$h_{B}=\frac{v^{2}}{2 g_{B}}$ .... (iii)
Dividing equation (ii) by equation (iii), we get
$\frac{h_{A}}{h_{B}}=\frac{g_{B}}{g_{A}}$
From equation (i), $g_{A}=9g_{B}$
$\frac{h_{A}}{h_{B}}=\frac{g_{B}}{9 g_{A}}=\frac{1}{9}\Rightarrow h_{B}=9h_{A}=9\times 2=18 \, m$
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