A ball is thrown vertically upward with a speed v. An identical second ball is thrown upward with a speed of 1/2 v (half as fast). What is the ratio of the maximum height of the second ball to that of the first ball

1:4

Explanation:

The formula for calculating the maximum height of the ball is expressed as;

H = u²sin²theta/2g

If a ball is thrown vertically upward with a speed v, its maximum height is expressed as;

H = v²sin²90/2g

Since sin90 = 1

H = v²/2g 1

If an identical second ball is thrown upward with a speed of 1/2 v, then the maximum height will be expressed as;

H1 = (1/2v)²sin²90/2g

H1 = (v²/4)(1)/2g

H1 = v²/8g 2

Divide both equations

H/H1 = (v²/2g)/(v²/8g)

H/H1 = v²/2g × 8g/v²

H/H1 = 8g/2g

H/H1 = 4

Reciprocate both sides

H1/H = 1/4

H1:H = 1:4

Hence the ratio of the maximum height of the second ball to that of the first ball is 1:4

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