Projectile Motion — Range, height, time of flight
Kinematics
13
JEE Qs
8%
Hard
75
min
Always resolve the initial velocity into horizontal and vertical components and analyze the motion independently along these directions, remembering that time is the common link.
🧮 Key Formulas
✅ Key Points for JEE
- 1Treat horizontal and vertical motion independently. Horizontal motion has zero acceleration (constant velocity); vertical motion has constant downward acceleration `g`.
- 2At the highest point of its trajectory, the vertical component of velocity is zero (v_y = 0), while the horizontal component remains constant.
- 3The trajectory of a projectile is always a parabola, assuming negligible air resistance and constant gravity.
- 4For a given initial speed, the horizontal range is the same for complementary angles of projection (θ and 90°-θ).
- 5The time of flight is determined primarily by the vertical motion and depends on the net vertical displacement; it is symmetrical for ground-to-ground projection.
⚠️ Common Mistakes
- ✕Incorrectly applying equations across horizontal and vertical components, e.g., using `v_y` with `u_x` or `a_x` with `s_y`.
- ✕Forgetting that the acceleration due to gravity (`g`) always acts downwards, sometimes leading to incorrect signs in vertical motion equations.
- ✕Assuming the horizontal component of velocity changes, rather than remaining constant throughout the flight (assuming no air resistance).
- ✕Mixing up formulas for ground-to-ground projection with those for projection from a height or up/down an inclined plane without proper modification.
📝 Practice Questions
See allQ27.A ball having kinetic energy KE, is projected at an angle of 60∘ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight ? (1) (KE) (2) (KE) 8 2 (3) (KE) (4) (KE) 16 4
Q28.The velocity-time graph of an object moving along a straight line is shown in figure. What is the distance covered by the object between t = 0 to t = 4 s ? (1) 30 m (2) 11 m (3) 10 m (4) 13 m
Q50.A particle is projected at an angle of 30∘ from horizontal at a speed of 60 m/s. The height traversed by the particle in the first second is h0 and height traversed in the last second, before it reaches the maximum height, is h1 . The ratio h0 : h1 is _________ [Take, g = 10 m/s2 ]
Q43.The motion of an airplane is represented by velocity-time graph as shown below. The distance covered by airplane in the first 30.5 second is _______ km . (1) 12 (2) 3 (3) 6 (4) 9
Q27.The position vector of a moving body at any instant of time is given as →r = . The magnitude (5t2^i −5t^j)m and direction of velocity at t = 2 s is, (1) 5√15 m/s, making an angle of tan−1 4 with - ve (2) 5√15 m/s, making an angle of tan−1 4 with + ve Y axis X axis (3) 5√17 m/s, making an angle of tan−1 4 with + ve (4) 5√17 m/s, making an angle of tan−1 4 with - ve X axis Y axis
Q31.Two projectiles are fired with same initial speed from same point on ground at angles of (45∘−α) and (45∘+ α), respectively, with the horizontal direction. The ratio of their maximum heights attained is : (1) 1−tan α (2) 1−sin 2α 1+tan α 1+sin 2α (3) 1+sin 2α (4) 1+sin α 1−sin 2α 1−sin α
NCERT Chapters
- Class 11 Physics Ch 3: Motion in a Straight Line
- Class 11 Physics Ch 4: Motion in a Plane