A ball rolls off a ledge at 9 m/s and lands 2.4 s later. Enter its range in m, to one decimal place.
The 20 m/s throw at 30° leaves the hand, arcs over level ground, and lands. One question has been waiting since the topic began: how far away does it land?
Nothing new arrives in this lesson. The range falls to three calculations you have already practiced: resolve the launch velocity, find the flight time from the vertical motion, and let the horizontal component carry the ball for that time.
Resolve the launch velocity first. For the 20 m/s throw at 30°, the horizontal component = 20 × cos 30° ≈ 17.3 m/s, and the vertical component at launch = 20 × sin 30° = 10 m/s.
The vertical motion alone sets the flight time. The time to the top = 10 ÷ 9.8 ≈ 1.0 s.
Over level ground the fall retraces the rise, so the total time of flight = 2 × 1.0 s ≈ 2.0 s.
The horizontal component never changes in flight, so the landing distance is plain distance = speed × time.
range = horizontal component × total time of flight = 17.3 × 2.0 ≈ 35 m
A golf shot leaves the tee at 28 m/s, 35° above the horizontal, over level ground. Calculate its range.
Write down the values in the question:
launch speed = 28 m/s at 35° above the horizontal, over level ground
range = ?
Write down the equations:
horizontal component = launch speed × cos θ
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
range = horizontal component × total time of flight
Substitute in the values:
horizontal component = 28 × cos 35° = 22.94 m/s
vertical component at launch = 28 × sin 35° = 16.06 m/s
time to the top = 16.06 ÷ 9.8 = 1.64 s
total time of flight = 2 × 1.64 = 3.28 s
range = 22.94 × 3.28
Simplify:
range = 75.2 m
A water balloon is lobbed at 15 m/s, 48° above the horizontal, over level ground. Calculate its range.
Write down the values in the question:
launch speed = 15 m/s at 48° above the horizontal, over level ground
range = ?
Write down the equations:
horizontal component = launch speed × cos θ
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
range = horizontal component × total time of flight
Substitute in the values:
horizontal component = 15 × cos 48° = 10.04 m/s
vertical component at launch = 15 × sin 48° = 11.15 m/s
time to the top = 11.15 ÷ 9.8 = 1.14 s
total time of flight = 2 × 1.14 = 2.28 s
range = 10.04 × 2.28
Simplify:
range = 22.9 m
Now run the chain on a fresh case, one stage at a time.
A goal kick leaves the turf at 24 m/s, 40° above the horizontal, over level ground. Enter its total time of flight in seconds, to one decimal place.
A goal kick leaves the turf at 24 m/s, 40° above the horizontal, over level ground. Enter its range in meters, to one decimal place.
A football is thrown at 21 m/s, 55° above the horizontal, over level ground. Enter its range in meters, to one decimal place.
A baseball is hit at 33 m/s, 25° above the horizontal, over level ground. Enter its range in meters, to one decimal place.
A softball is thrown at 17 m/s, 62° above the horizontal, over level ground. Enter its range in meters, to one decimal place.
An arrow leaves a bow at 42 m/s, 18° above the horizontal, over level ground. Enter its range in meters, to one decimal place.