← Free fall

The rise and the fall take equal times

A ball is tossed straight up at 19.6 m/s. Air resistance is ignored. How long does it take to reach its highest point? Enter the time in seconds.

Toss a ball straight up and catch it at the height it left from. Which takes longer: the rise to the top, or the fall back down?

launch height24.5 m/sA ball leaves the launch height at 24.5 m/s, straight up. Air resistance ignored.
rising, one second between ticks24.5 m/st = 014.7 m/st = 1 s4.9 m/st = 2 sEach second of the rise removes 9.8 m/s of upward speed: 24.5, then 14.7, then 4.9 m/s.
launch heightthe top, t = 2.5 s2.5 s after launch, the rise ends: the ball is at the top — v = 0 for an instant, theacceleration still 9.8 m/s² downward.
falling, one second between ticks4.9 m/st = 3 s14.7 m/st = 4 s24.5 m/st = 5 slaunch heightEach second of the fall adds 9.8 m/s of downward speed: 4.9, then 14.7, then 24.5 m/s at thelaunch height.
time t (s)velocity v (m/s)2.55+24.50−24.52.5 s up2.5 s downThe whole flight at once: 2.5 s from the launch height to the top, 2.5 s from the top backto it.
An ideal up-and-down flight is symmetric: the rise and the fall between the same two heights take equal times.

The two times match because free-fall acceleration means the velocity changes by 9.8 m/s downward with each second of flight. Each second of the rise removes 9.8 m/s of upward speed. Each second of the fall adds 9.8 m/s of downward speed.

'Ideal' means air resistance is ignored. With air resistance the two times differ, so the rule is stated for ideal flights.

Try it on a fresh case. A cricket ball thrown straight up takes 3.5 s to reach the top of its flight, with air resistance ignored. The fall from the top back to the height it was thrown from takes 3.5 s too: the rise and the fall between the same two heights take equal times.

A beanbag tossed straight up rises for 1.8 s before reaching the top of its flight. Air resistance is ignored. How long does the fall from the top back to the launch height take? Enter the time in seconds.

A ball tossed straight up rises for 2.2 s to the top of its flight. Air resistance is ignored. How long after launch does it pass back through its launch height? Enter the time in seconds.

A juggling club is tossed straight up and later caught at the height it left from. Air resistance is ignored. Compare the time the club spends rising with the time it spends falling.

Your turn

A ball is at the highest point of its flight 3.2 s after leaving the thrower's hand. Air resistance is ignored. At what time after launch does it pass back through the height it was released from? Enter the time in seconds.

Dana watches a tossed orange rise and fall back to hand height and claims the fall was quicker than the rise, 'because the orange speeds up as it falls.' Air resistance is ignored. Which response is correct?

A ball thrown straight up passes a windowsill 0.6 s before reaching the top of its flight. Air resistance is ignored. How long after the top does it pass the windowsill again on the way down? Enter the time in seconds.