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## Homework Statement

A ball of mass 65 g is thrown vertically upwards from ground level with a speed of 16 m s–1. Air resistance is negligible. The ball takes time t to reach maximum height. For time t 2 after the ball has been thrown, calculate the ratio of the potential energy of ball to the kinetic energy of ball .

## Homework Equations

T=(usinA)/g, where A=90

s= 1/2at^2 + ut

GPE=mgh

KE=1/2mv^2

## The Attempt at a Solution

T=(usinA)/g

= 16/9.81

= 1.63s

so t/2= 0.816

at 0.816s,

s= (1/2 x 9.81 x 0.816^2) + (16 x 0.816)

= 16.322m

GPE=mgh

0.065 x 9.81 x 16.322

KE= 1/2mv^2

1/2 x 0.065 x v^2

v^2-u^2=2as

v^2=2 x 9.81 x 16.322

substituting,

KE= 1/2 x 0.065 x 2 x 9.81 x 16.322

= 0.065 x 9.81 x 16.322

So, GPE:KE

= (0.065 x 9.81 x 16.322) / (0.065 x 9.81 x 16.322)

=1:1

The answer is incorrect. The correct answer is 3:1.

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