A ball is thrown at an upward angle over a field. At the very top of its arc, which way is the ball moving?
A ball leaves a thrower's hand at 20 m/s, 30° above the horizontal, and climbs toward the top of its arc. How long does the climb take? You have timed climbs like this before, but only for throws that went straight up.
For a straight-up throw, you have already calculated the time to the top: set v = 0 at the top in v = v_0 + a × t, and with up positive the time comes out as the launch velocity divided by 9.8. The question is what to feed that routine when the launch is angled.
An angled launch is not a straight-up throw. But its vertical motion is still ordinary free fall, running on the vertical component alone.
So the straight-up routine still works, on one condition: feed it the vertical component of the launch velocity, never the full launch speed.
For the 20 m/s throw at 30°, the vertical component is 20 × sin 30° = 10 m/s.
time to the top = 10 ÷ 9.8 ≈ 1.0 s
Careful with the one mistake this routine invites: 20 ÷ 9.8 ≈ 2.0 s uses the full launch speed and comes out twice too long here. Only the vertical component climbs.
A baseball leaves the bat at 24 m/s, 30° above the horizontal. Calculate the time to the top of its flight.
Write down the values in the question:
launch speed = 24 m/s at 30° above the horizontal
time to the top = ?
Write down the equations:
vertical component at launch = launch speed × sin θ
time to the top = vertical component at launch ÷ 9.8
Substitute in the values:
vertical component at launch = 24 × sin 30° = 12 m/s
time to the top = 12 ÷ 9.8
Simplify:
time to the top = 1.2 s
A snowball is thrown at 18 m/s, 56° above the horizontal. Calculate the time to the top of its flight.
Write down the values in the question:
launch speed = 18 m/s at 56° above the horizontal
time to the top = ?
Write down the equations:
vertical component at launch = launch speed × sin θ
time to the top = vertical component at launch ÷ 9.8
Substitute in the values:
vertical component at launch = 18 × sin 56° = 14.92 m/s
time to the top = 14.92 ÷ 9.8
Simplify:
time to the top = 1.5 s
Now run the routine on a fresh case, one step at a time.
A football is punted at 16 m/s, 35° above the horizontal. Enter the vertical component of its launch velocity in m/s, to one decimal place.
A football is punted at 16 m/s, 35° above the horizontal. Enter the time to the top of its flight in seconds, to one decimal place.
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.
A tennis ball is hit at 26 m/s, 42° above the horizontal. Enter its time to the top in seconds, to one decimal place.
A volleyball leaves a bump at 14 m/s, 65° above the horizontal. Enter its time to the top in seconds, to one decimal place.
A lacrosse ball is thrown at 40 m/s, 33° above the horizontal. Enter its time to the top in seconds, to one decimal place.