MHT-CET Physics · Motion in a Plane

Projectiles

A projectile moves steadily at u cos θ horizontally while its vertical motion is free fall starting at u sin θ; that gives time of flight 2u sin θ/g, greatest height u²sin²θ/2g and range u²sin 2θ/g on level ground.

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Why this matters

6 PYQs, one HARD: time of flight, the angle at which height equals range, the kinetic energy and the radius of curvature at the top, the height reached in terms of two times, and a ball thrown from a tower. One card.

Concept 1 of 1: Time of Flight, Height and Range

Split the motion: horizontally nothing acts, so uₓ = u cos θ stays; vertically it is free fall from u sin θ. At the top only the horizontal velocity remains, so the kinetic energy there is E cos²θ, and gravity acts as the centripetal force on a path of radius u²cos²θ/g. Setting H = R gives tan θ = 4. A ball thrown upward that passes height h at t₁ on the way up and lands t₂ later has h = ½gt₁t₂. From a tower, solve the vertical equation for the time first, taking care whether the throw is above or below the horizontal, then multiply by the horizontal speed.

Definition

$T = \dfrac{2u\sin\theta}{g}$ , $H = \dfrac{u^2\sin^2\theta}{2g}$ , $R = \dfrac{u^2\sin 2\theta}{g}$ (196 m/s at 30° ⇒ T = 20 s).
At the top: $KE = E\cos^2\theta$ ; radius of curvature $\dfrac{u^2\cos^2\theta}{g}$ .
H = R ⇒ $\tan\theta = 4$ .
Vertical throw passing h at $t_1$ , landing $t_2$ later: $h = \tfrac{1}{2}gt_1t_2$ .
From height h: solve $-h = u_yt - \tfrac{1}{2}gt^2$ (u_y negative if thrown below the horizontal).

Projectile

T = \frac{2u\sin\theta}{g}, \qquad H = \frac{u^2\sin^2\theta}{2g}, \qquad R = \frac{u^2\sin 2\theta}{g}

Worked example

A ball is thrown horizontally at 15 m/s from a 20 m cliff. How far out does it land? (g = 10 m/s²)

Practice this conceptself-check · 1 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q5Moderate

Example 1 · Motion in a Plane · Projectile Motion — Range, Height, and Time of Flight

For a projectile, the maximum height and horizontal range are same. The angle of projection

\theta

of the projectile is

Using cos θ for the vertical motion

The vertical component is u sin θ; the horizontal is u cos θ. Time of flight and height come from sin θ.

Assuming a throw from a tower goes upward

'At 30° with the horizontal' does not say which side. A throw 30° below lands 8.7 m out from a 10 m tower at 10 m/s; 30° above lands 17.3 m out.

Drill 5 more on time of flight, height and range

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Time of Flight, Height and Range
Projectile
$T = \frac{2u\sin\theta}{g}, \qquad H = \frac{u^2\sin^2\theta}{2g}, \qquad R = \frac{u^2\sin 2\theta}{g}$

Watch out for (2)

Using cos θ for the vertical motion
→ Time of Flight, Height and Range
Assuming a throw from a tower goes upward
→ Time of Flight, Height and Range

Test yourself on Motion in a Plane

15 past MHT-CET questions from this chapter, timed at 14 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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