JEE Physics Projectile Motion MCQs Set 03

Answer: (b) \( u\cos\theta \)

Explanation:
1. At the top the vertical component is zero and only the unchanged horizontal component \( u\cos\theta \) remains.

Teacher's Note:
1. The horizontal velocity of a projectile stays constant.
2. The speed is minimum at the highest point.

Answer: (a) \( 10\sqrt7 \) m/s

Explanation:
1. \( v_x = 30\cos30^\circ = 15\sqrt3 \) and \( v_y = 30\sin30^\circ-10(1) = 5 \).
2. \( v = \sqrt{675+25} = \sqrt{700} = 10\sqrt7 \) m/s.

Teacher's Note:
1. Resolve into components, apply gravity to the vertical one only.
2. Combine with Pythagoras.

Answer: (b) \( 20\sqrt3 \) m/s, \( 60^\circ \)

Explanation:
1. The top is reached at \( t = 3 \) s, so \( u_y = 10\times3 = 30 \) m/s.
2. At \( t = 2 \) s, \( v_y = 10 \) and \( \tan30^\circ = \frac{10}{u_x} \), so \( u_x = 10\sqrt3 \) m/s.
3. \( u = \sqrt{900+300} = 20\sqrt3 \) m/s and \( \tan\theta = \frac{30}{10\sqrt3} = \sqrt3 \), so \( \theta = 60^\circ \).

Teacher's Note:
1. Horizontal travel at the top gives the time to the peak.
2. The horizontal component is constant throughout.

Answer: (d) \( 20\sqrt2 \) m/s

Explanation:
1. The top is at height \( 20\sqrt2\sin45^\circ = 20 \) m. It leaves the top at \( 45^\circ \) with speed v, and the range 40 m needs \( \frac{v^2}{g} = 40 \), so \( v^2 = 400 \) (g = 10).
2. Energy conservation: \( u^2 = v^2+2gh = 400+400 = 800 \), so \( u = 20\sqrt2 \) m/s.

Teacher's Note:
1. Use energy conservation on the smooth slope and the range formula for the jump.
2. The body leaves the incline at an angle equal to the incline angle.

Answer: (c) \( b = 2a \)

Explanation:
1. \( R = \frac{2ab}{g} \) and \( H = \frac{b^2}{2g} \).
2. \( R = 2H \) gives \( \frac{2ab}{g} = \frac{b^2}{g} \), so \( b = 2a \).

Teacher's Note:
1. Express R and H in terms of the velocity components.
2. The condition R = 2H corresponds to \( \tan\theta = 2 \).

Answer: (c) 5 m

Explanation:
1. The time to the top is 1 s, so \( H = \frac12gt^2 = 5 \) m.

Teacher's Note:
1. The time to rise equals the time to fall for a level landing.
2. Treat the descent as free fall from rest.

Answer: (c) \( \sqrt2 \) times

Explanation:
1. \( R\propto u^2 \) and \( T\propto u \) at a fixed angle.
2. Doubling R multiplies u by \( \sqrt2 \), so T becomes \( \sqrt2 \) times.

Teacher's Note:
1. Range scales as the square of the speed, time of flight as the speed.
2. Eliminate u between the two relations.

Answer: (c) \( \tan\alpha:1 \)

Explanation:
1. First: \( T_1 = \frac{2u\sin\alpha}{g} \). Second makes \( 90^\circ-\alpha \) with the horizontal: \( T_2 = \frac{2u\cos\alpha}{g} \).
2. \( T_1:T_2 = \tan\alpha:1 \).

Teacher's Note:
1. An angle from the vertical is the complement of the angle from the horizontal.
2. Time of flight is proportional to the sine of the angle with the horizontal.

Answer: (b) The direction in which he leaps and the initial speed

Explanation:
1. Range \( = \frac{u^2\sin2\theta}{g} \) depends on the take-off speed and the angle only.
2. Mass does not enter, so option (b) is correct.

Teacher's Note:
1. Projectile range is independent of the mass.
2. Only speed and angle of take-off matter.

Answer: (c) 32 m/s

Explanation:
1. Maximum range \( = \frac{u^2}{g} \), so \( u = \sqrt{10\times100} = 31.6\approx32 \) m/s.

Teacher's Note:
1. The range is maximum at a launch angle of 45 degrees.
2. Then range equals \( u^2/g \).

Answer: (a) Their ranges are the same

Explanation:
1. Range is \( \frac{u^2\sin2\theta}{g} \), and \( \sin60^\circ = \sin120^\circ \), so the ranges are equal.
2. Heights and times of flight depend on \( \sin\theta \), which differs.

Teacher's Note:
1. Complementary angles of projection give equal ranges.
2. Height and flight time differ for the two angles.

Answer: (c) \( 1:1 \)

Explanation:
1. By energy conservation both land with \( v^2 = u^2+2gh \), whatever the direction of projection.
2. The ratio is 1 : 1.

Teacher's Note:
1. Energy conservation: the final speed depends only on the initial speed and the height through which it drops.
2. Direction does not matter.

Answer: (b) 20 s

Explanation:
1. The vertical motion is free fall: \( t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times1960}{9.8}} = 20 \) s.

Teacher's Note:
1. The horizontal velocity does not affect the fall time.
2. Use h = ½gt² for the vertical motion.

Answer: (b) \( \frac{500}{3} \) m

Explanation:
1. \( t = \sqrt{\frac{2\times490}{9.8}} = 10 \) s and \( v = 60\) km/h \( = \frac{50}{3} \) m/s.
2. Distance \( = \frac{50}{3}\times10 = \frac{500}{3} \) m.

Teacher's Note:
1. Convert km/h to m/s by multiplying by 5/18.
2. The bomb keeps the horizontal speed of the plane.

Answer: (a) The range of the projectile is less than the initial distance between the gun and the target

Explanation:
1. Both fall the same distance \( \frac12gt^2 \) below the line of sight, so they meet if the bullet reaches the target's horizontal position before the ground.
2. If the range is less than the distance, the bullet hits the ground first and misses.

Teacher's Note:
1. The bullet and the target have the same vertical acceleration.
2. A shot that cannot cover the horizontal distance fails regardless of the drop of the target.

Answer: (c) 18 m/s

Explanation:
1. At \( 45^\circ \), \( \tan45^\circ = \frac{v_y}{v_x} = 1 \), so \( v_y = v_x = 18 \) m/s.

Teacher's Note:
1. The horizontal component stays 18 m/s.
2. Equal components correspond to 45 degrees.

Answer: (a) \( 30^\circ \)

Explanation:
1. \( R_{up} = \frac{u^2}{g(1+\sin\alpha)} \) and \( R_{down} = \frac{u^2}{g(1-\sin\alpha)} \).
2. \( \frac{1+\sin\alpha}{1-\sin\alpha} = 3 \) gives \( \sin\alpha = \frac12 \), so \( \alpha = 30^\circ \).

Teacher's Note:
1. Use the maximum range formulas for projection along an incline.
2. The up-slope range is smaller because gravity has a component against the motion.

Answer: (c) 30 m

Explanation:
1. The angle with the slope is \( 30^\circ \). Range along the incline \( = \frac{2v^2\sin(\beta-\alpha)\cos\beta}{g\cos^2\alpha} \).
2. \( = \frac{2\times441\times\frac12\times\frac12}{9.8\times\frac34} = 30 \) m.

Teacher's Note:
1. Resolve motion along and perpendicular to the incline.
2. The formula uses the angle of the barrel with the horizontal.

Answer: (c) 4 km

Explanation:
1. \( R_{max} = \frac{u^2}{g} = 6 \) km and \( R_{up,max} = \frac{u^2}{g(1+\sin30^\circ)} = \frac{6}{1.5} = 4 \) km.

Teacher's Note:
1. The maximum range up an incline of angle \( \alpha \) is \( \frac{u^2}{g(1+\sin\alpha)} \).
2. Set \( \alpha = 0 \) to recover the horizontal formula.

Answer: (c) The horizontal component of momentum will be conserved

Explanation:
1. No horizontal force acts, so the horizontal momentum is conserved.
2. At the top the ball still has horizontal speed (KE not zero), gravity changes the vertical momentum, and the potential energy is maximum at the top.

Teacher's Note:
1. Momentum is conserved only in the direction with no net force.
2. Gravity acts vertically, so only the horizontal component is conserved.