Horizontal and vertical components of projectile motion, range, maximum height, time of flight.
Answer at least 3 of 3 correctly to complete this section.
A footballer kicks a ball at an angle. A stone is thrown from a cliff. A dart is launched towards a board. Every time an object moves through the air under gravity alone, it follows a parabolic path — and we can predict exactly where it will land.
Projectile motion combines everything you have learnt about SUVAT with the idea of resolving vectors into components. The beautiful trick is that the horizontal and vertical motions are completely independent — solve them separately, and the variable t (time) connects them.
A projectile launched at speed u at angle θ above the horizontal has:
Once launched (ignoring air resistance):
The two directions are independent. We apply SUVAT to each direction separately.
At 45° from ground level, the range is maximised for this speed.
Since a=0 horizontally:
x=(ucosθ)⋅t
That is it — constant velocity, so distance = speed × time.
Taking upwards as positive, with a=−g:
vy=usinθ−gt
y=(usinθ)t−21gt2
vy2=(usinθ)2−2gy
A ball is kicked from ground level at 20 m s−1 at 30° above the horizontal. Find the horizontal and vertical components of the initial velocity.
ux=20cos30°=20×23=103≈17.3 m s−1
uy=20sin30°=20×0.5=10 m s−1
At the highest point: The vertical velocity is zero (vy=0), but the horizontal velocity is still ucosθ. The projectile is NOT stationary at the top — it is moving horizontally.
Answer at least 3 of 3 correctly to complete this section.
If a projectile is launched from and lands at the same height, then the vertical displacement is y=0 when it lands.
y=(usinθ)t−21gt2=0
t(usinθ−21gt)=0
So t=0 (launch) or:
T=g2usinθ
At the highest point, vy=0:
0=(usinθ)2−2gH
H=2gu2sin2θ
The range is the horizontal distance when t=T:
R=(ucosθ)⋅T=ucosθ⋅g2usinθ=gu2⋅2sinθcosθ
Using the double angle formula 2sinθcosθ=sin2θ:
R=gu2sin2θ
Maximum range: Since sin2θ is greatest when 2θ=90°, i.e. θ=45°, the maximum range for a given launch speed occurs at 45°.
A ball is kicked from ground level at 24.5 m s−1 at 30° above the horizontal. Find (a) the time of flight, (b) the maximum height, (c) the range.
(a) Time of flight:
T=9.82×24.5×sin30°=9.82×24.5×0.5=9.824.5=2.5 s
(b) Maximum height:
H=2×9.824.52×sin230°=19.6600.25×0.25=19.6150.0625≈7.66 m (3 s.f.)
(c) Range:
R=9.824.52×sin60°=9.8600.25×23=9.8600.25×0.8660≈53.0 m (3 s.f.)
A projectile is launched at 30 m s−1. At what two angles will it achieve a range of 80 m?
R=gu2sin2θ
80=9.8900sin2θ
sin2θ=90080×9.8=900784=0.8711
2θ=60.6°or2θ=119.4°
θ=30.3°orθ=59.7°
Complementary angles: For any given range (less than the maximum), there are always two launch angles that achieve it. They add up to 90°. The flatter trajectory gets there faster; the steeper one goes higher.
Answer at least 3 of 3 correctly to complete this section.
The trajectory equation gives y in terms of x (eliminating t). This describes the shape of the path.
From the horizontal equation: x=(ucosθ)t, so t=ucosθx.
Substitute into the vertical equation:
y=(usinθ)⋅ucosθx−21g(ucosθx)2
y=xtanθ−2u2cos2θgx2
This is a quadratic in x — confirming the path is a parabola.
A projectile is launched at 20 m s−1 at 45° from ground level. Find the equation of the trajectory.
y=xtan45°−2(400)cos245°9.8x2
y=x−2×400×0.59.8x2
y=x−4009.8x2
y=x−0.0245x2
This equation tells us the height at any horizontal distance. Setting y=0:
0=x(1−0.0245x)
x=0orx=0.02451≈40.8 m
The range is approximately 40.8 m. We can verify: R=gu2sin90°=9.8400≈40.8 m. It matches.
The trajectory equation is particularly useful when:
A ball is kicked at 25 m s−1 at 40° towards a wall 30 m away. The wall is 5 m high. Does the ball clear the wall?
Using the trajectory equation:
y=30tan40°−2×625×cos240°9.8×302
y=30×0.8391−2×625×0.58689.8×900
y=25.17−733.58820
y=25.17−12.03=13.14 m
Since 13.14>5, the ball clears the wall by over 8 m.
Answer at least 3 of 3 correctly to complete this section.
A stone is thrown horizontally at 12 m s−1 from the top of a cliff 40 m above sea level. Find (a) the time taken to reach the sea, (b) the horizontal distance from the base of the cliff, (c) the speed at which it hits the sea.
(a) Thrown horizontally means θ=0°, so the initial vertical velocity is zero.
Taking downwards as positive for the vertical direction:
s=ut+21gt2
40=0+21(9.8)t2
t2=9.880=8.163
t=2.857≈2.86 s (3 s.f.)
(b) Horizontal distance:
x=12×2.857=34.3 m (3 s.f.)
(c) At impact, find both components of velocity:
Speed = magnitude of velocity vector:
v=vx2+vy2=144+784=928=30.5 m s−1 (3 s.f.)
The direction below the horizontal is:
α=arctan(1228.0)=66.8°
A ball is thrown from the top of a building 25 m high at 15 m s−1 at 40° above the horizontal. Find the time taken for the ball to reach the ground.
Taking upwards as positive, the ground is at y=−25:
uy=15sin40°=9.642 m s−1
y=uyt−21gt2
−25=9.642t−4.9t2
4.9t2−9.642t−25=0
Using the quadratic formula:
t=9.89.642±93.17+490
t=9.89.642±583.17
t=9.89.642±24.15
Taking the positive root:
t=9.89.642+24.15=9.833.79=3.45 s (3 s.f.)
Answer at least 5 of 6 correctly to complete this section.
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