The 20 m/s throw at 30° climbs for about 1.0 s. How high is it at the top? For straight-up throws you have answered this before. The angled version turns on the same choice as the time to the top: which speed do you feed in?
For a straight-up throw, you have already calculated the maximum height: set v = 0 at the top in v² = v_0² + 2 × a × Δx, and with up positive the height comes out as the launch velocity squared divided by (2 × 9.8). The climb of an angled launch runs on its vertical component alone.
The straight-up routine also works for an angled launch, on the same condition as the time to the top: 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.
maximum height = 10² ÷ (2 × 9.8) ≈ 5.1 m
The same mistake lurks here: 20² ÷ (2 × 9.8) ≈ 20.4 m uses the full launch speed and lands four times too high for this throw. Only the vertical component climbs.
Maximum height is measured above the launch point.
A softball leaves the bat at 28 m/s, 30° above the horizontal. Calculate its maximum height above the launch point.
Write down the values in the question:
launch speed = 28 m/s at 30° above the horizontal
maximum height = ?
Write down the equations:
vertical component at launch = launch speed × sin θ
maximum height = (vertical component at launch)² ÷ (2 × 9.8)
Substitute in the values:
vertical component at launch = 28 × sin 30° = 14 m/s
maximum height = 14² ÷ (2 × 9.8) = 196 ÷ 19.6
Simplify:
maximum height = 10.0 m
A volleyball is set at 16 m/s, 68° above the horizontal. Calculate its maximum height above the launch point.
Write down the values in the question:
launch speed = 16 m/s at 68° above the horizontal
maximum height = ?
Write down the equations:
vertical component at launch = launch speed × sin θ
maximum height = (vertical component at launch)² ÷ (2 × 9.8)
Substitute in the values:
vertical component at launch = 16 × sin 68° = 14.83 m/s
maximum height = 14.83² ÷ (2 × 9.8) = 219.93 ÷ 19.6
Simplify:
maximum height = 11.2 m
Now run the routine on a fresh case, one step at a time.
A tennis lob leaves the racket at 22 m/s, 35° above the horizontal. Enter the vertical component of its launch velocity in m/s, to one decimal place.
A tennis lob leaves the racket at 22 m/s, 35° above the horizontal. Enter its maximum height above the launch point in meters, to one decimal place.
A kickball is booted at 18 m/s, 52° above the horizontal. Enter its maximum height above the launch point in meters, to one decimal place.
A golf drive leaves the tee at 32 m/s, 25° above the horizontal. Enter its maximum height above the launch point in meters, to one decimal place.
A baseball line drive leaves the bat at 45 m/s, 20° above the horizontal. Enter its maximum height above the launch point in meters, to one decimal place.
At a fall fair, a pumpkin launcher fires at 27 m/s, 58° above the horizontal. Enter the pumpkin's maximum height above the launch point in meters, to one decimal place.