A 4.03-kg object is thrown vertically upward from the surface of Mercury, where

Question

A 4.03-kg object is thrown vertically upward from the surface of Mercury, where the acceleration due to gravity is g1= 3.70 m/s/s. The initial velocity is v1, and the object reaches a maximum height of y1. What is the maximum height, y2, if the object is thrown with a speed of v2=1.15v1 from the surface of the Dwarf planet Pluto? The acceleration due to gravity on Pluto is g2= 0.658 m/s/s. Give your answer as a multiple of y1.

The answer box it wants the answer in looks like this

y2=         x y1

(the x is a multiplication symbol and the answer fits imbetween the = and the x)

Explanation / Answer

y=-.5*g*t2+v*t

v=-g*t+vo

At the maximum height, the velocity is 0 m/s. Therefore, we know the time for the object to reach the top on Mercury is:

t=(v1/g)*s=(v1/3.7) s

which means it will reach a height of

y1=-.5*(v12/3.7)*m+v12/3.7 m=(v12/7.4)m

If the object was thrown on pluto, with a velocity of 1.15 that on mercury, the time would change to:

t=(v1/g)*s=(v1/.658) s

meaning the height would change to:

y2=-.5*(v12/.658)*m+v12/.658 m=(v12/1.316)m

If you divide y2 by y1, you get the ratio which is:

(v12/1.316)m/(v12/7.4)m=7.4/1.316=5.62

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