Generated Example: A ball is thrown straight upward from a point 1.2 m above the ground. After 3.0 s, the ball returns to the ground. Determine the maximum height reached. Use g = 9.81 m/s².

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0–50 m
0.5–10 s

Motion model

+s upward1.2 m: releasemaximum heightrelease
Position1.20 m
Velocity14.32 m/s
Acceleration−9.81 m/s²

Guided solution

1

Establish signs and states

Choose upward as positive. Therefore gravity is negative: a = −9.81 m/s².

2

Find the unknown initial velocity

m/s
3

Recognize the top condition

At maximum height the ball changes direction, so its instantaneous velocity is:

m/s
4

Find maximum height without time

m

Final result

Why solve for v₀ first?

The total flight time and final position are known, but the launch velocity is not. The position-time equation connects those known values directly.

Why is velocity zero at the top?

The ball pauses for an instant while changing from upward to downward motion. Acceleration is still −9.81 m/s² there; only velocity is zero.

Engineering Record

Explain the coordinate sign convention and why the velocity is zero at maximum height while acceleration is not.