Gravity on a lonely planet

Classical Mechanics Level 1

Astronauts on an atmosphere-less planet wanted to find the acceleration due to gravity. In order to find it, they launched a ball from the surface of the planet, with a velocity of 15 m/s , 15\text{ m/s}, and predicted that the ball would reach a height of s = 15 t − 1 2 g p t 2 s=15t-\dfrac 12 g_pt^2 meters t t seconds after the launch. If the ball reached the maximum height after 20 20 seconds, what was the value of g p ? g_p?

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The answer is 0.75.

Sravanth C. by Sravanth C. , 21, India

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2 solutions

Sravanth C.
Mar 2, 2016

We know that velocity is the derivative of the displacement function. Thus; v ( t ) = d s d t = d d s ( 15 t − 1 2 g p t 2 ) = 15 − g p t v(t)=\dfrac{ds}{dt}=\dfrac{d}{ds}(15t-\dfrac 12 g_pt^2)\\=15-g_pt

And we know at the maximum height, the velocity of the ball will be 0 m / s 0m/s . So; 15 − g p t = 0 15 − g p × 20 = 0 ∴ g p = 15 20 = 0.75 15-g_pt=0\\15-g_p\times 20=0\\\therefore g_p= \dfrac{15}{20}=0.75

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You can directly use u/g to find the time of ascent(time taken to reach the maximum height) = 15/g =20

Abhiram Rao - 5 years, 1 month ago
Nitin Kumar
Nov 10, 2019

s=ut+0.5at^2 but s=15t-0.5gt^2 so comparing, we get u=15 and a=-g. it is given that t=20s when v=0=u-at=15-20g so g=0.75m/s^2

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