Question

The value of $g$ on the moon is ${\dfrac{1}{6}^{th}}$ of the value of $g$ on earth. A man can jump $1.5m$ high on the earth. He can jump on the moon up to a height of:
A. $9m$
B. $7.5m$
C. $6m$
D. $4.5m$

Answer 1

Hint Velocity of man on the highest point is zero and we know the height at which man can jump on earth hence we can calculate the initial velocity at which a man jumps on earth. Now the initial velocity of the man remains the same on the moon. Hence using the equation of motion, we can calculate the height to which man can jump on the moon.

Complete Step by step solution
For earth:
Given,
$\Rightarrow$ $a = - g$ (for upward motion)

$\Rightarrow$ $v = 0$ (at highest point)

$\Rightarrow$ $h = 15m$
Now from third equation of motion, we get
${v^2} = {u^2} + 2as$
On putting values, we get
$0 = {u^2} - 2g(1.5) \\\ 0 = {u^2} - 3 \times 10 \\\ {u^2} = 30 \\\ u = \sqrt {30} ......(1) \\\$
For moon:
Given,
$\Rightarrow$ $a = \dfrac{{ - g}}{6}$
$\Rightarrow$ $v = 0$
and, from (1) we get $u = \sqrt {30} m{s^{ - 1}}$
hence using the equation ${v^2} = {u^2} + 2as$ , we get
$0 = 30 - 2 \times \dfrac{g}{6} \times H \\\ H = \dfrac{{180}}{{2g}} \\\ H = 9m \\\$
Hence, man can jump to a height of 9m on moon.

Option (A) is correct.

Note We had considered the motion of man as a straight-line motion. Since the equation we had used above is valid only in case of straight-line motion and for constant acceleration. This straight-line motion is also called Linear motion or rectilinear motion.