Question

A ball is shot from the ground straight up into the air with initial velocity of 40 ft/sec. Assuming that the air resistance can be ignored, how high does it go

Answer 1

The height to which the ball attained, given the data is 7.58 m

Data obtained from the question

• Initial velocity (u) = 40 ft/s = 40 × 0.3048 = 12.192 m/s
• Final velocity (v) = 0 m/s (at maximum height)
• Acceleration due to gravity (g) = 9.8 m/s²
• Maximum height (h) =?

How to determine the maximum height

v² = u² – 2gh (since the ball is going against gravity)

0² = 12.192² – (2 × 9.8 × h)

0 = 12.192² – 19.6h

Collect like terms

0 – 12.192² = –19.6h

–12.192² = –19.6h

Divide both side by –19.6

h = –12.192² / –19.6

h = 7.58 m

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