A golf ball is chipped at 30 m/s, 50° above the horizontal. Enter its time to the top in seconds, to one decimal place.
The 20 m/s throw at 30° reaches the top of its arc at about 1.0 s. The top is not the end of the story. The ball still has to come down. How long is the whole flight?
You have already seen the time symmetry of an ideal up-and-down flight: the rise and the fall between the same two heights take equal times. That symmetry is about to do all the work.
A projectile's vertical motion is an ideal up-and-down flight.
Over level ground, the landing height matches the launch height. So the fall from the top retraces the rise: same two heights, equal times.
The 20 m/s throw at 30° reaches its top at about 1.0 s, so its total time of flight = 2 × 1.0 s = 2.0 s.
A thrown ball reaches the top of its arc 1.8 s after launch and lands back at its launch height. How long is it in the air in total?
Write down the values in the question:
time to the top = 1.8 s, landing height = launch height
total time of flight = ?
Write down the equation:
total time of flight = 2 × time to the top
Substitute in the values:
total time of flight = 2 × 1.8
Simplify:
total time of flight = 3.6 s
A punted football reaches the top of its arc 1.4 s after launch and lands back at launch height. Enter its total time of flight in seconds.
A tossed tennis ball returns to its launch height. Why is its total flight time double its time to the top?
The doubling is legal only over level ground, where the landing height matches the launch height.
Land on a rooftop and the fall is cut short. Land in a ditch and it runs long. In both cases the total is no longer double the time to the top.
A ball is kicked at 23 m/s, 41° above the horizontal, over level ground. Calculate its total time of flight.
Write down the values in the question:
launch speed = 23 m/s at 41° above the horizontal, over level ground
total time of flight = ?
Write down the equations:
vertical component at launch = launch speed × sin θ
time to the top = vertical component at launch ÷ 9.8
total time of flight = 2 × time to the top
Substitute in the values:
vertical component at launch = 23 × sin 41° = 15.09 m/s
time to the top = 15.09 ÷ 9.8 = 1.54 s
total time of flight = 2 × 1.54
Simplify:
total time of flight = 3.1 s
A soccer ball is kicked at 26 m/s, 50° above the horizontal, over level ground. Enter its total time of flight in seconds, to one decimal place.
A softball is thrown at 19 m/s, 42° above the horizontal, over level ground. Enter its total time of flight in seconds, to one decimal place.
Three balls are thrown upward at an angle, air resistance ignored. One lands back at its launch height, one lands on a deck above its launch height, one lands in a dip below its launch height. For which ball is the total flight time exactly double the time to the top?
A home-run ball leaves the bat at 31 m/s, 36° above the horizontal, over level ground. Enter its total time of flight in seconds, to one decimal place.