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PhysicsEasyClass 11

Projectile Motion

Kinematics

42

JEE Qs

7%

Hard

90

min

Always resolve the initial velocity into horizontal and vertical components and analyze motion independently in these two perpendicular directions using 1D kinematic equations.

🧮 Key Formulas

Vx = u cos(theta)
Vy = u sin(theta) - gt
x = (u cos(theta)) * t
y = (u sin(theta)) * t - (1/2)gt^2
T = (2u sin(theta)) / g
H = (u^2 sin^2(theta)) / (2g)
R = (u^2 sin(2*theta)) / g
y = x tan(theta) - (gx^2) / (2u^2 cos^2(theta))

✅ Key Points for JEE

  • 1Projectile motion is a combination of two independent motions: uniform velocity horizontally and uniformly accelerated motion vertically under gravity.
  • 2Horizontal component of velocity (Vx) remains constant throughout the motion, as horizontal acceleration is zero.
  • 3Vertical component of velocity (Vy) changes due to constant downward acceleration 'g'. At the highest point, Vy = 0.
  • 4The trajectory is symmetric about the highest point: time to reach maximum height equals time to fall from maximum height to the same horizontal level.
  • 5For a given launch speed, the horizontal range is maximum when the projection angle is 45 degrees, and ranges are equal for complementary angles (theta and 90-theta).

⚠️ Common Mistakes

  • Confusing horizontal and vertical components of velocity and acceleration, or applying vertical acceleration 'g' to horizontal motion.
  • Incorrectly using sign conventions for displacement, velocity, or acceleration due to gravity, especially when motion is upwards then downwards.
  • Attempting to solve problems without resolving initial velocity into its horizontal and vertical components first.

📝 Practice Questions

See all

Q27.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

2025·MCQEasy

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

2025·Graph basedEasy

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 ]

2025·NumericalMedium

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

2025·NumericalEasy

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

2025·MCQMedium

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

2025·MCQEasy

NCERT Chapters

  • Class 11 Physics Ch 4: Motion in a Plane