Question

An ISRO astronaut projects a ball vertically upwards with same velocity from the surface of Earth and the surface of
Moon. It takes time t1 to reach maximum height on the Earth while t2 time to reach maximum height on the Moon. If
gravity on Moon is one-sixth to that on Earth, find the ratio of t2/t1. (Neglect air friction)

Answer 1

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$


$\underline{\textsf{Let the ball was projected with the velocity of u.}}$


$\underline{\textsf{On Earth : }} \\ \\ \sf \implies Initial \: Velocity \: = \: u \\ \\ \sf \implies Final \: Velocity \: = \: 0 \\ \\ \sf \implies Time \: = \: t_1 \\ \\ \sf \implies Acceleration \: due \: to \: gravity \: = \: -g \\ \\ \sf \implies Maximum \: Height \: = \: h_1$



$\textsf{Now,} \\ \\ \sf \implies v \: = \: u \: - \: gt \\ \\ \sf \implies 0 \: = \: u \: -gt_1 \\ \\ \sf \: \: \therefore \: \: u \: = \: g t_1 \qquad...(1)$





$\underline{\textsf{On Moon : }} \\ \\ \sf \implies Initial \: Velocity \: = \: u \\ \\ \sf \implies Final \: Velocity \: = \: 0 \\ \\ \sf \implies Time \: = \: t_2 \\ \\ \sf \implies Acceleration \: due \: to \: gravity \: = \: \dfrac{ - g}{6} \\ \\ \sf \implies Maximum \: Height \: = \: h_2$




$\textsf{Now,} \\ \\ \sf \implies v \: = \: u \: - \: gt \\ \\ \sf \implies 0 \: = \: u \: - \: \dfrac{g}{ 6} t_2 \\ \\ \sf \: \: \therefore \: \: u \: = \: \dfrac{g}{6} t_2 \qquad...(2)$



$\underline{\textsf{Divide (1) by (2),}} \\ \\ \sf \implies \dfrac{u}{u} \: = \: gt_1 \: \div \: \dfrac{g}{6}t_2 \\ \\ \sf \implies 1 \: = \: gt_1 \: \times \: \dfrac{6}{gt_2} \\ \\ \sf \implies 1 \: = \: \dfrac{6t_1}{t_2} \\ \\ \sf \: \: \therefore \: \: \dfrac{t_2}{t_1} \: = \: 6$

Answer 2

Heya...

Ques :-

An ISRO astronaut projects a ball vertically upwards with same velocity from the surface of Earth and the surface of
Moon. It takes time t1 to reach maximum height on the Earth while t2 time to reach maximum height on the Moon. If
gravity on Moon is one-sixth to that on Earth, find the ratio of t2/t1. (Neglect air friction)???

Ans ____________

On earth :-

Initial velocity = u
Final velocity = v
Acceleration due to gravity = -g
Time taken = t1

Equation :- v = u - gt
Now....

v = u-gt1 .......
u = gt1........1

On moon :-

lnitialVelocity = u
Final velocity = v
time taken = t2
acceleration = -g/6

Equation :- v = u-gt
Now....
v = u-g/6t2
u = g/6t2 .....2


By dividing both the equation 1&2

u/u = gt1/gt2

1 = 6t1/t2

= 6 Ans ..

Thank you