- Topic
- Antiderivatives
- Difficulty
- Standard
- Suggested time
- 5 min
- Marks
- 3
- Theory
- CLP-1 §4.1Key results
Antiderivatives practice question 6
A ball is thrown straight up from a height of 1 m with an initial velocity of 14.7 m/s. Its acceleration is a constant m/s². Find the maximum height it reaches.
Give the answer in metres.
At the top, the ball stops rising for an instant. What is its velocity then, and how do you get from the acceleration back to the height?
Antidifferentiate twice: acceleration to velocity (using ), then velocity to position (using ).
and . Find when and evaluate there.
A decimal.
Theory CLP-1 §4.1Key results
Worked solution
Strategy
Constant acceleration integrates to linear velocity and quadratic position, with the two initial conditions fixing the two constants. The peak is where velocity is zero.
Working
Antiderivative of acceleration.
Antiderivative of velocity.
Peak height.
Answer
The answer is 12.025 metres.
Check
The speed falls steadily from 14.7 m/s to 0 in 1.5 s, averaging 7.35 m/s, so it climbs 7.35 × 1.5 = 11.025 m above its 1 m start.
Where this goes wrong
Dropping one of the constants of integration, typically the starting height of 1 m, which gives 11.025. Each antidifferentiation needs its own initial condition.
Leans on
Velocity and acceleration, from The definition of the derivative
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