Projectile Motion — Find Maximum Height Reached

medium JEE-MAIN NEET • JEE Main 2024 • 4 min read
Tags Kinematics

Maximum height in projectile motion is one of the most straightforward formulas to apply, but students often get confused about which component of velocity matters. Here’s the key: at the highest point, the vertical component of velocity becomes zero. The horizontal component continues unchanged. Use this insight and the rest follows naturally.

Question

A ball is projected at an angle of 30° with the horizontal with a speed of 20 m/s. Find:

  1. The maximum height reached
  2. The speed of the ball at maximum height
  3. The time taken to reach maximum height

(Take g=10g = 10 m/s²)

Solution — Step by Step

Step 1: Resolve initial velocity into components.

ux=ucos⁡30°=20×32=103≈17.3 m/su_x = u\cos 30° = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \approx 17.3 \text{ m/s} uy=usin⁡30°=20×0.5=10 m/su_y = u\sin 30° = 20 \times 0.5 = 10 \text{ m/s}

Step 2: Find maximum height.

At maximum height, vertical velocity =0= 0. Using vy2=uy2−2gHv_y^2 = u_y^2 - 2gH:

0=(10)2−2×10×H0 = (10)^2 - 2 \times 10 \times H H=10020=5 mH = \frac{100}{20} = 5 \text{ m}

Or directly using the formula:

H=u2sin⁡2θ2g=400×sin⁡230°2×10=400×0.2520=5 mH = \frac{u^2\sin^2\theta}{2g} = \frac{400 \times \sin^2 30°}{2 \times 10} = \frac{400 \times 0.25}{20} = 5 \text{ m}

Step 3: Find speed at maximum height.

At maximum height, vertical velocity =0= 0. Only horizontal velocity remains:

v=ux=ucos⁡30°=103≈17.3 m/sv = u_x = u\cos 30° = 10\sqrt{3} \approx 17.3 \text{ m/s}

Step 4: Find time to reach maximum height.

vy=uy−gtv_y = u_y - gt 0=10−10t0 = 10 - 10t t=1 st = 1 \text{ s}

Or using Ttotal=2usin⁡θg=2×20×0.510=2T_{total} = \dfrac{2u\sin\theta}{g} = \dfrac{2 \times 20 \times 0.5}{10} = 2 s, so time to max height =T2=1= \dfrac{T}{2} = 1 s.

Answers:

  • Maximum height =5= \mathbf{5} m
  • Speed at maximum height =103= 10\sqrt{3} m/s ≈\approx 17.3 m/s
  • Time to reach maximum height == 1 second

Why This Works

Projectile motion is two independent motions happening simultaneously:

  • Horizontal: No force, so constant velocity ucos⁡θu\cos\theta throughout.
  • Vertical: Gravity decelerates the upward motion. Vertical velocity decreases from usin⁡θu\sin\theta to 0 at the top, then increases downward.

Maximum height is reached when all the vertical kinetic energy is converted to gravitational potential energy. That’s why we set vy=0v_y = 0 — it marks the exact moment the ball stops rising and starts falling.

The speed at maximum height is not zero. Only the vertical component is zero. The ball is still moving horizontally at ucos⁡θu\cos\theta. This is a very commonly tested conceptual point in NEET.

Alternative Method — Energy Method

Using conservation of energy for the vertical motion:

Initial vertical KE =12muy2=12m(usin⁡θ)2= \dfrac{1}{2}m u_y^2 = \dfrac{1}{2}m(u\sin\theta)^2

This equals the gain in PE at maximum height: mgHmgH

mgH=12m(usin⁡θ)2mgH = \frac{1}{2}m(u\sin\theta)^2 H=u2sin⁡2θ2g=400×0.2520=5 mH = \frac{u^2\sin^2\theta}{2g} = \frac{400 \times 0.25}{20} = 5 \text{ m}

The energy method is quicker and avoids dealing with the velocity equation — very useful when you just need HH.

Important Results to Memorize

At θ=30°\theta = 30°: H=u28gH = \dfrac{u^2}{8g}

At θ=45°\theta = 45°: H=u24gH = \dfrac{u^2}{4g} (also equals Rmax⁡/4R_{\max}/4)

At θ=60°\theta = 60°: H=3u28gH = \dfrac{3u^2}{8g}

At θ=90°\theta = 90° (vertical throw): H=u22gH = \dfrac{u^2}{2g} (maximum possible height)

Note: H60°=3×H30°H_{60°} = 3 \times H_{30°} for the same initial speed.

Common Mistake

Using the full speed uu instead of the vertical component usin⁡θu\sin\theta in the formula. The height depends only on the vertical motion. Writing H=u22g=40020=20H = \dfrac{u^2}{2g} = \dfrac{400}{20} = 20 m uses the full velocity — this is only correct for a vertical throw (θ=90°\theta = 90°). For any angle θ\theta, you must use usin⁡θu\sin\theta as the effective initial vertical velocity.

JEE Main often gives this as a two-part problem: find max height, then find the range. Don’t recalculate everything from scratch for the range. You already have uu, θ\theta, and gg — just apply R=u2sin⁡2θgR = \dfrac{u^2\sin 2\theta}{g} directly. For θ=30°\theta = 30°: R=400×sin⁡60°10=400×0.86610≈34.6R = \dfrac{400 \times \sin 60°}{10} = \dfrac{400 \times 0.866}{10} \approx 34.6 m.

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