Projectile Motion and Kinematics

A projectile is an object moving after launch under gravity in the ideal model. We neglect air resistance and take gravity as constant and downward.

A thrown cricket ball and a launched cannonball are familiar examples. Their real paths can differ from the ideal model because air exerts forces.

Kinematics describes motion using position, velocity and acceleration. Projectile motion applies that description to horizontal and vertical components.

Take the launch point as the origin. Choose rightwards as positive x and upwards as positive y.

Let u be launch speed and θ the angle above the horizontal.

• uₓ = u cos θ: initial horizontal velocity.
• uᵧ = u sin θ: initial vertical velocity.
• g: the positive magnitude of gravitational acceleration.

The horizontal and vertical components describe the same object’s motion. They share the same elapsed time.

Why analyse them separately? Gravity acts vertically in this model. It changes vertical velocity while leaving horizontal velocity unchanged.

For motion along one axis with constant acceleration:

• v = u + at
• s = ut + ½at²
• v² = u² + 2as

Here, u and v are initial and final signed velocities. s is signed displacement, rather than total distance travelled.

Acceleration a is measured in m/s². Time t is measured in seconds, and displacement in metres.

Use these equations separately for each axis. Along the horizontal axis, aₓ = 0. With upwards positive, vertical acceleration is aᵧ = −g.

Near Earth’s surface, g ≈ 9.8 m/s² is a common approximation. Use 10 m/s² only when the question permits that rounding.

In the ideal model, no horizontal force acts after release. Horizontal acceleration is therefore zero.

The horizontal velocity remains vₓ = u cos θ. Horizontal displacement after time t is x = uₓt.

Equal time intervals give equal horizontal displacements. The ball does not need a continuing forward push to retain its horizontal velocity.

With upwards positive, vᵧ = uᵧ − gt. Vertical displacement is y = uᵧt − ½gt².

The positive vertical velocity decreases while the projectile rises. At the peak, vᵧ = 0.

During descent, vertical velocity becomes negative. Its downward speed increases as gravity continues to act.

Zero vertical velocity does not mean zero acceleration. The acceleration remains downward throughout the ideal flight.

An angled projectile normally retains horizontal velocity at the peak. Its total speed there is |uₓ|, because the vertical component is zero.

A straight vertical throw has no horizontal velocity. Its total speed becomes zero briefly at the highest point.

Both still have downward acceleration. Distinguish zero vertical velocity from zero total velocity.

Time to peak. Set vᵧ = 0 in vᵧ = uᵧ − gt.

tₚₑₐₖ = uᵧ/g = u sin θ/g.

The formula uses the initial vertical component, rather than the full launch speed for every angle.

A projectile with nonzero horizontal velocity follows a parabolic path under constant vertical gravity and negligible air resistance.

From x = uₓt, substitute t = x/uₓ into the vertical displacement equation.

y = x tan θ − gx²/(2u² cos² θ).

The squared x term produces the parabola. A straight vertical throw is a special case with no horizontal trajectory span.

The ideal path is symmetric about the vertical line through its peak. The visible flight ends where the projectile meets the landing surface.

Different launch and landing heights can leave unequal rising and falling portions. Real air resistance can also alter the ideal shape.

For a projectile returning to its launch height, set y = 0.

0 = t(uᵧ − ½gt).

One solution is t = 0, the launch instant. The later solution gives the flight duration:

T = 2uᵧ/g = 2u sin θ/g.

Ascent and descent to the original height take equal times in this model. The total is therefore twice the time to peak.

Landing above or below the launch height changes that duration. Solve the vertical displacement equation for the actual landing level.

Maximum height H is the vertical rise above the launch point. It is not automatically the height above the ground.

Use the vertical equation vᵧ² = uᵧ² − 2gy. The vertical velocity vᵧ is zero at the peak.

H = uᵧ²/(2g) = u² sin² θ/(2g).

An elevated launch point adds its initial height to that rise when calculating peak height above the ground. Keep the reference level explicit.

Range is the horizontal distance from launch to landing. The familiar formula below applies when landing occurs at the launch height.

Multiply horizontal velocity by the same-height flight duration:

R = uₓT = (u cos θ)(2u sin θ/g).

Using 2 sin θ cos θ = sin 2θ gives:

R = u² sin 2θ/g.

This is horizontal distance, rather than the curved distance travelled along the path. For unequal landing height, calculate flight time from vertical motion and then use R = uₓT.

At fixed launch speed and g, R = u² sin 2θ/g. The angle changes the sin 2θ factor.

That factor reaches its maximum value of 1 when 2θ = 90°. The launch angle is therefore 45°.

The result requires negligible air resistance and equal launch and landing heights. An elevated launch or aerodynamic forces can change the best angle.

Two complementary launch angles add to 90°. At fixed speed and g, they give equal same-height ranges in the ideal model.

For 30°, sin 2θ = sin 60°. For 60°, sin 2θ = sin 120°. These sine values are equal.

Their flight times and maximum heights differ. The steeper 60° launch has a larger vertical component.

Equal range does not mean an identical trajectory. The 30° path is lower and takes less time.

Consider a launch at 20 m/s and 30°, with g = 10 m/s². Ignore air resistance and let the ball land at its launch height.

Split the velocity. The vertical component is 20 sin 30° = 10 m/s. The horizontal component is 20 cos 30° ≈ 17.32 m/s.

Find the times. Time to peak is 10/10 = 1 s. Total flight time is 2 s.

Find maximum height. H = 10²/(2 × 10) = 5 m.

Find range. R ≈ 17.32 × 2 = 34.64 m.

At the peak, vertical velocity is zero while horizontal velocity remains about 17.32 m/s. These quantities describe one flight, using the same time and assumptions.

Compare two objects starting at the same height and the same instant. One is dropped from rest; the other is launched horizontally.

Both have zero initial vertical velocity. With identical g and negligible air resistance, their vertical motions match.

They therefore reach the same horizontal ground level together. The horizontally launched object also travels sideways.

Worked example. Let the launch height be 20 m, with g = 10 m/s².

Using downward distance, 20 = ½ × 10 × t². The positive time is t = 2 s.

The dropped object has the same falling time. A different initial vertical velocity would change that comparison.

The National Defence Academy and Naval Academy (NDA & NA) General Ability Test (GAT) uses familiar kinematics calculations.

In NDA & NA II 2016, Series A Q73, a ball is thrown vertically upwards at 25.2 m/s. This solution uses g = 9.8 m/s² to find the time to reach the peak.

At the peak, vertical velocity is zero. Therefore:

t = u/g = 25.2/9.8 ≈ 2.57 s.

The question does not specify g. Using g = 10 would give 2.52 s instead. Do not confuse time to peak with total flight time back to the original height.

Full NDA & NA II 2016 GAT question.