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Projectile launched at an angle

SPH3U Grade 11 Physics · Kinematics

A school soccer match doubles as a data-collection afternoon for the kinematics unit. The goalkeeper sets the ball at the edge of the goal area and punts it off the turf toward the far end of the field, while two students work the sideline: one aims the radar gun the team uses at practice, the other films the launch against the painted lines of the grandstand. The radar gun clocks the ball leaving the boot at 86.4 km/h, and a protractor laid over a freeze-frame of the video puts the launch at 38.0° above the horizontal. The ball climbs, hangs for a moment at the top of its arc, and comes down on level ground near the far penalty area. Ignoring air resistance, find how long the ball stays airborne, how high it climbs at the top of its arc, the horizontal speed it carries the whole way, and how far downfield it lands.

86.4 km/h off the boot38.0°apex 11.1 mrange 57.0 m3.01 s in the air

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • v₀ = 86.4 km/h — Launch speed off the boot
  • θ = 38 ° — Launch angle above the horizontal
Determine
  1. (a)how long the ball stays airborne
  2. (b)how high it climbs at the top of its arc
  3. (c)the horizontal speed it carries the whole way
  4. (d)how far downfield it lands
Step 1 of 4(a) · solve for Time of flight

The gun reads km/h and every kinematic equation below wants m/s, so 86.4 ÷ 3.6 = 24.0 m/s happens before anything else. Hang time then depends only on the vertical part of that launch, and it is the clock everything downstream runs on.

θv0T
Rearranged for T
T=2v0sin⁡θgT = \frac{2 v_0 \sin\theta}{g}
Your values, in your units
T=2 (86.4 km/h) sin⁡(38 ∘)(9.80665 m/s2)T = \frac{2 \, \left( 86.4\ \text{km/h} \right) \, \sin \left( 38\ ^{\circ} \right)}{\left( 9.80665\ \text{m/s}^{2} \right)}
Converted to base units
T=2 (24 m/s) sin⁡(38 ∘)(9.80665 m/s2)T = \frac{2 \, \left( 24\ \text{m/s} \right) \, \sin \left( 38\ ^{\circ} \right)}{\left( 9.80665\ \text{m/s}^{2} \right)}
Answer
T=3.0134 sT = 3.0134\ \text{s}

Carried onward at full precision, not this rounded figure.

Open the Projectile Time of Flight solver →

Step 2 of 4(b) · solve for Maximum height

The apex comes from the same two givens. Nothing later in the chain needs it — it is asked because clearance over a crossbar or a treeline is a real question.

θv0H
Rearranged for H
H=v02sin⁡2θ2gH = \frac{v_0^{2} \sin^{2}\theta}{2g}
Your values, in your units
H=(86.4 km/h)2 sin⁡2(38 ∘)2(9.80665 m/s2)H = \frac{\left( 86.4\ \text{km/h} \right)^{2} \, \sin^{2} \left( 38\ ^{\circ} \right)}{2\left( 9.80665\ \text{m/s}^{2} \right)}
Converted to base units
H=(24 m/s)2 sin⁡2(38 ∘)2(9.80665 m/s2)H = \frac{\left( 24\ \text{m/s} \right)^{2} \, \sin^{2} \left( 38\ ^{\circ} \right)}{2\left( 9.80665\ \text{m/s}^{2} \right)}
Answer
H=11.132 mH = 11.132\ \text{m}

Carried onward at full precision, not this rounded figure.

Open the Projectile Maximum Height solver →

Step 3 of 4(c) · solve for Horizontal velocity

Now the horizontal slice. With no air to slow it, this 18.9 m/s — about 68 km/h — is the same at the boot as it is at the grass.

θvvx
Rearranged for vx
vx=vcos⁡θv_x = v \cos\theta
Your values, in your units
vx=(86.4 km/h) cos⁡(38 ∘)v_x = \left( 86.4\ \text{km/h} \right) \, \cos \left( 38\ ^{\circ} \right)
Converted to base units
vx=(24 m/s) cos⁡(38 ∘)v_x = \left( 24\ \text{m/s} \right) \, \cos \left( 38\ ^{\circ} \right)
Answer
vx=18.912 m/sv_x = 18.912\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Horizontal Velocity Component solver →

Step 4 of 4(d) · solve for Displacement

Range is horizontal speed times time aloft. Because vₓ never changes, the initial and final velocities in this average are the same carried number from step 3, running for the clock set in step 1.

v0vtd
Rearranged for d
d=v0+v2 td = \frac{v_0 + v}{2} \, t
18.912 m/scarried from step 3
18.912 m/scarried from step 3
3.0134 scarried from step 1
Your values, in your units
d=(18.9123 m/s)+(18.9123 m/s)2 (3.01344 s)d = \frac{\left( 18.9123\ \text{m/s} \right) + \left( 18.9123\ \text{m/s} \right)}{2} \, \left( 3.01344\ \text{s} \right)
Answer
d=56.991 md = 56.991\ \text{m}

Carried onward at full precision, not this rounded figure.

Open the Displacement from Average Velocity solver →

Answer

Therefore the ball hangs in the air for 3.01 s, tops out 11.1 m above the turf, carries a constant 18.9 m/s of horizontal speed, and comes down 57.0 m downfield — deep in the far penalty area.

Why this order

The order matters more than the algebra. Time of flight goes first because it is the one quantity both halves of the motion share: gravity fixes it, and the horizontal motion simply runs for that long. A student who reaches straight for the packaged range formula R = v₀² sin 2θ ⁄ g gets the right number and learns nothing, then stalls the moment the ball lands on a roof instead of level ground. Building the range as vₓ × T survives that change; the shortcut does not.

Three errors are near-universal. The first happens before any physics does: 86.4 km/h is not 86.4 of anything the equations recognise, and a launch speed left in km/h quietly inflates every answer by 3.6. The second is swapping sine and cosine — anchor it on the extremes, since a launch straight up has no horizontal component and cos 90° = 0 says so. The third is treating the doubled angle in the range formula as the launch angle. Underneath all of it sits Galileo's independence principle from Two New Sciences (1638): the horizontal and vertical motions never talk to each other, which is why the 38° in step 1 and the 38° in step 3 do completely different jobs with the same number.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.