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Q.What is a projectile? Explain the five factors that affect the trajectory of a projectile in sports. Write the range formula for level ground, and use it to calculate and compare the range of an object projected at 20 m/s at

(i) 30° and
(ii) 45°, taking g = 9.8 m/s². Comment on why the optimum angle in shot put, javelin and long jump is not 45°.
Sikkim CbseNCERTLong· 5mImportance★★★★★est
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A projectile is any object launched into the air that moves under the influence of gravity alone. Its trajectory is shaped by five key factors: angle of projection, speed of projection, height of projection, gravity, and air resistance. The range formula for level ground is R = (v² sin 2θ)/g, and the optimum angle in throwing events is less than 45° because the release point is above the landing surface.

What is a Projectile?

A projectile is any object that is thrown, hit, kicked, or otherwise launched into the air and then moves under the sole influence of gravity (and air resistance, in real conditions). Once it leaves the thrower's hand or the implement, the only force acting on it — ignoring air — is its own weight. This gives it a curved path called a trajectory, which in a vacuum is a perfect parabola.

In sports, projectiles are everywhere: a basketball shot, a javelin, a shot put, a long jumper's body, a soccer ball in flight, a cricket ball after being bowled. Understanding what controls that flight path is the difference between a good throw and a gold medal.

Five Factors Affecting the Trajectory

1. Angle of Projection (θ)

This is the angle at which the projectile is launched relative to the horizontal. It determines the balance between vertical and horizontal components of the initial velocity. A steeper angle gives more height but less horizontal speed; a shallower angle gives more horizontal speed but less flight time. The range is maximised at 45° when launch and landing heights are equal.

2. Speed (Velocity) of Projection (v)

The magnitude of the initial velocity. This is the single most important factor — range is proportional to v². Doubling the release speed quadruples the range. In events like javelin or shot put, athletes train to maximise this speed through technique and strength.

3. Height of Projection (Release Height)

The vertical distance between the release point and the landing surface. In most throwing events, the projectile is released from above the ground (e.g., shoulder height). This effectively increases the range compared to a launch from ground level, and it shifts the optimum angle downward — because the extra height means the projectile has more time to travel horizontally before hitting the ground.

4. Gravity (g)

Gravity pulls the projectile downward at 9.8 m/s² on Earth. It is constant for all projectiles near the surface. A lower gravity (e.g., on the Moon) would give a much longer range; a higher gravity would shorten it. In sports, we cannot change gravity, but we must account for it.

5. Air Resistance (Drag) and Spin

Air resistance opposes motion, slowing the projectile and reducing range. Its effect depends on the projectile's shape, cross-sectional area, and speed. Spin (like backspin on a basketball or topspin on a tennis ball) creates a Magnus force that can curve the trajectory. In events like javelin, aerodynamic design is critical.

Important

For exam purposes, the five factors are: angle, speed, height, gravity, and air resistance. Always list them in this order.

Range Formula for Level Ground

When a projectile is launched from and lands at the same height, with no air resistance, the horizontal range is given by:

R = (v² sin 2θ)/g

Where:

  • R = horizontal range (metres)
  • v = initial speed (m/s)
  • θ = angle of projection (degrees)
  • g = acceleration due to gravity (9.8 m/s²)

The sin 2θ term is key: it reaches its maximum value of 1 when 2θ = 90^°, i.e., θ = 45^°. So for level ground, 45° gives the greatest possible range.

Calculation: Comparing 30° and 45°

Given v = 20 m/s, g = 9.8 m/s².

(i) At θ = 30^°:

  1. Compute sin 2θ = sin 60^° = √(3)/2 ≈ 0.8660
  2. R = (20² × 0.8660)/9.8 = (400 × 0.8660)/9.8 = 346.4/9.8 ≈ 35.35 m

(ii) At θ = 45^°:

  1. Compute sin 2θ = sin 90^° = 1
  2. R = (20² × 1)/9.8 = 400/9.8 ≈ 40.82 m

So the range at 45° (40.82 m) is greater than at 30° (35.35 m), as expected. …

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