Answer:
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
