Projectile motion numericals: Class 11 practice
A projectile launched at speed u and angle θ has time of flight T = 2u sin θ / g, maximum height H = u² sin²θ / 2g and horizontal range R = u² sin 2θ / g (level ground, no air resistance).. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Equations of motion
- sin and cos of 30°, 45°, 60°
The idea behind projectile motion numericals
A projectile launched at speed u and angle θ has time of flight T = 2u sin θ / g, maximum height H = u² sin²θ / 2g and horizontal range R = u² sin 2θ / g (level ground, no air resistance).
Step-by-step method
- Split u into u cos θ (horizontal) and u sin θ (vertical).
- Pick the formula for T, H or R.
- Take g as the question states (10 m/s² here).
Worked example 1 (Basic)
A ball is projected at 50 m/s at 45° to the horizontal. Find its time of flight (g = 10 m/s²).
- T = 2u sin θ / g = 2 × 50 × sin 45° / 10
- = 7.07 s
- Answer: 7.07 s
Worked example 2 (Standard)
A ball is projected at 10 m/s at 30° to the horizontal. Find its maximum height (g = 10 m/s²).
- H = u² sin²θ / 2g = 100 × sin²30° / 20
- = 1.25 m
- Answer: 1.25 m
Worked example 3 (Challenge)
A ball is projected at 10 m/s at 30° to the horizontal. Find its horizontal range (g = 10 m/s²).
- R = u² sin 2θ / g = 100 × sin 60° / 10
- = 8.66 m
- Answer: 8.66 m
Common mistakes and misconceptions
- Using sin θ instead of sin 2θ for range
- Forgetting to square sin θ for height
Practice questions
- (Basic) A ball is projected at 20 m/s at 30° to the horizontal. Find its time of flight (g = 10 m/s²).
- (Basic) A ball is projected at 20 m/s at 45° to the horizontal. Find its time of flight (g = 10 m/s²).
- (Standard) A ball is projected at 10 m/s at 60° to the horizontal. Find its maximum height (g = 10 m/s²).
- (Standard) A ball is projected at 30 m/s at 30° to the horizontal. Find its maximum height (g = 10 m/s²).
- (Challenge) A ball is projected at 50 m/s at 45° to the horizontal. Find its horizontal range (g = 10 m/s²).
- (Challenge) A ball is projected at 50 m/s at 60° to the horizontal. Find its horizontal range (g = 10 m/s²).
Answer key with full working
- Q1: T = 2u sin θ / g = 2 × 20 × sin 30° / 10 → = 2.00 s → Answer: 2.00 s
- Q2: T = 2u sin θ / g = 2 × 20 × sin 45° / 10 → = 2.83 s → Answer: 2.83 s
- Q3: H = u² sin²θ / 2g = 100 × sin²60° / 20 → = 3.75 m → Answer: 3.75 m
- Q4: H = u² sin²θ / 2g = 900 × sin²30° / 20 → = 11.25 m → Answer: 11.25 m
- Q5: R = u² sin 2θ / g = 2500 × sin 90° / 10 → = 250.00 m → Answer: 250.00 m
- Q6: R = u² sin 2θ / g = 2500 × sin 120° / 10 → = 216.51 m → Answer: 216.51 m
Syllabus note
This skill is listed under "Motion in a Plane" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is projectile motion numericals in?
"Motion in a Plane", Class 11 Physics (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.