How to solve Projectile Velocity at Maximum Height in NEET — Step-by-step

Numerical/Problem-solving

1. TITLE

How to solve Projectile Velocity at Maximum Height in NEET — Step-by-step

2. PROBLEM TYPE INTRO

This category of numerical is based on the 2D kinematics of projectile motion. We will need the horizontal velocity component formula: $v_x = v_0 \cos\theta_0$. The most typical mistake students make is assuming that a projectile's total velocity is zero at its maximum height, completely forgetting that the horizontal component remains constant throughout the flight.

3. MCQ (Full numerical problem)

Question:
A cricket ball is launched with an initial velocity of $20 \text{ m s}^{-1}$ at an angle of $60^{\circ}$ above the horizontal. Neglecting air resistance, what is the magnitude of its velocity exactly at its maximum height?

(a) $0 \text{ m s}^{-1}$
(b) $10 \text{ m s}^{-1}$
(c) $17.3 \text{ m s}^{-1}$
(d) $20 \text{ m s}^{-1}$

4. STEP-BY-STEP SOLUTION

SOLUTION
Given:
Initial velocity ($v_0$) = $20 \text{ m s}^{-1}$
Angle of projection ($\theta_0$) = $60^{\circ}$

To Find:
The magnitude of the velocity of the projectile at its maximum height.

Formula:
$v_x = v_0 \cos\theta_0$
Use this formula because the horizontal component of a projectile's velocity remains constant throughout its flight (since horizontal acceleration $a_x = 0$). At the maximum height, the vertical component of the velocity ($v_y$) becomes momentarily zero, leaving only the horizontal component.

Step 1: Identify the velocity components at maximum height.
At maximum height, $v_y = 0 \text{ m s}^{-1}$.
The total velocity is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{v_x^2 + 0^2} = v_x$.

Step 2: Calculate the horizontal velocity component.
$v_x = v_0 \cos\theta_0$
$v_x = 20 \cos(60^{\circ})$

Step 3: Solve for the final value.
$v_x = 20 \times 0.5$
$v_x = 10 \text{ m s}^{-1}$

Answer: $10 \text{ m s}^{-1}$ (Option b)

5. WHERE STUDENTS GO WRONG

List 2-3 common mistakes:
- ❌ Mistake 1: Assuming total velocity is $0 \text{ m s}^{-1}$ at maximum height. Only the vertical velocity ($v_y$) is zero; the horizontal velocity ($v_x$) is non-zero (unless launched straight up at $90^\circ$).
- ❌ Mistake 2: Using the wrong trigonometric component by calculating $v_y = v_0 \sin\theta_0$ instead. This gives the initial vertical velocity (which would result in $17.3 \text{ m s}^{-1}$), not the velocity at the peak.
- ❌ Mistake 3: Thinking the velocity remains the initial $20 \text{ m s}^{-1}$ throughout the trajectory. The acceleration due to gravity changes the vertical speed continuously.

6. QUICK FORMULA REFERENCE

Formulas Used:

$v_x = v_0 \cos\theta_0$ — Use when finding the constant horizontal velocity of a projectile at any point during its flight.
$v_y = v_0 \sin\theta_0 - g t$ — Use when calculating the changing vertical velocity of a projectile at a given time $t$.

7. SIMILAR PROBLEM HINT

"If NEET changes the values, the method stays the same. Practice with: A football is kicked at $30 \text{ m s}^{-1}$ at an angle of $30^{\circ}$. Find its velocity at maximum height."

8. NCERT REFERENCE

Based on: NCERT Class 11, Chapter 3, Example 3.8

9. CTA

More solved numericals → Kinematics Complete Guide
Practice problems → Mock exams