Projectile Motion — Angle for maximum range
Kinematics
13
JEE Qs
8%
Hard
45
min
Always check if the projectile lands on a horizontal surface at the same height as the launch point before assuming 45° for maximum range.
🧮 Key Formulas
✅ Key Points for JEE
- 1For projectile motion over a horizontal plane, the maximum range (R_max) is achieved when the angle of projection θ = 45°.
- 2This condition arises because the range formula R = (u^2 * sin(2θ)) / g is maximized when sin(2θ) = 1, which implies 2θ = 90° or θ = 45°.
- 3The maximum range possible for a given initial speed u is R_max = u^2 / g.
- 4The range of a projectile is the same for complementary angles of projection, i.e., R(θ) = R(90° - θ).
- 5The angle of 45° for maximum range is strictly valid only when the projectile is launched and lands at the same horizontal level.
⚠️ Common Mistakes
- ✕Assuming 45° is *always* the angle for maximum range, even when the launch and landing points are at different heights or on an inclined plane.
- ✕Confusing the angle for maximum range (45°) with the angle for maximum height (90°).
- ✕Incorrectly maximizing sin(θ) instead of sin(2θ) when deriving the condition for maximum range.
📝 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 4: Motion in a Plane