Question

A body is thrown upward from earths surface with velocity 5m/s and from planet surface with velocity 3m/s. Both follow the same path. What is the projectile acc. Due to gravity on the planet?

Answer 1

Answer:

Explanation:

v^2 =u^2 +2as

At max height v=0 and for upward direction a=-g

therefore, u^2 =2gs

s=u^2/2g;

since, se=sp

(ue/up)^2=(ge/gp)

(5/3)^2 =9.8/gp

gp=3.5m/s

Answer 2

Given that,

Initial velocity from the surface of Earth, u = 5 m/s

Initial velocity from the surface of planet, u' = 3 m/s

To find,

Acceleration due to gravity on the planet.

Solution,

Acceleration in earth's surface is g while for planet it is g'. At maximum height, final velocity is equal to 0. Using third equation of motion,

$v^2-u^2=2as$

s = height reached by object in both surfaces

a = -g

$s=\dfrac{u^2}{2g}$

s is same for both. So,

$\dfrac{u^2}{2g}=\dfrac{u'^2}{2g'}\\\\\dfrac{u^2}{g}=\dfrac{u'^2}{g'}\\\\g'=\dfrac{u'^2g}{u^2}\\\\g'=\dfrac{(3)^2\times 9.8}{(5)^2}\\\\g'=3.52\ m/s^2$

So, the acceleration due to gravity on the planet is $3.52\ m/s^2$.

Learn more,

Acceleration

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