Important Questions on Projectile Motion | Physics for EmSAT Achieve PDF Download

Important Questions on Projectile Motion

Q1: In a normal projectile motion, what will be the condition for maximum range?
(a) θ = 45°
(b) θ = 30°
(c) θ = 60°
(d) θ = 0°
Ans: (a)
R = v²(sin 2θ)/g is the formula for horizontal range. Hence, at sin 2θ = 1, the value of R will be maximum, which indicates that 2θ = 90°, this means that θ should be 45°. Hence the correct answer is θ = 45°.

Q2: In a normal projectile motion, what will be the condition for maximum height?
(a) θ = 55°
(b) θ = 90°
(c) θ = 180°
(d) θ = 0°
Ans: (b)
h = (v sinθ)²/2g is the formula for height. Hence, at sin θ = 1, this will be maximum, which means that θ should be 90°. Hence the correct answer will be θ = 90°.

Q3: In a normal projectile motion, which of the following quantities has an effect in calculating the body’s mass?
(a) Force
(b) Time
(c) Velocity
(d) Horizontal range
Ans: (a)
All the quantities mentioned above are kinematic quantities, except force. Since Force = m x a is a kinetic quantity, the force is affected by the mass of the body.

Q4: In a normal projectile motion, what will be the angle of the projectile to get the horizontal range minimum?
(a) θ = 90°
(b) θ = 30°
(c) θ = 45°
(d) θ = 75°
Ans: (a)
R = v² (sin 2θ)/g is the formula for horizontal range when θ is taken as 90°, sin 2θ will be sin 180° = 0. Hence, the range covered will be 0, the minimum value it can achieve.

Q5: An object of mass 2000 g covers a maximum vertical distance of 6 m when it is projected at an angle of 45° from the ground. Calculate the velocity with which it was thrown. Take g = 10 m/s²,
(a) 12.10 m/s
(b) 15.49 m/s
(c) 2.155 m/s
(d) 12.0 m/s
Ans: (a)
Here, h = 6, θ = 45°, g = 10 and h = (v sinθ) ²/2g is the formula for maximum height. Solving the equation by substituting the values, we get the initial velocity as 15.49 m/s.

Q6: A bullet shot hits the ground 4 km away, which was at an angle of 30° with the plane. What should be the projection angle to hit a target 6 km away? Neglect air resistance and assume fixed muzzle speed.
Solution:
Given, Range, R=4 km
Angle of projection = 30°
g=9.8m/s²
For the projection velocity u₀ , the horizontal range is given by:
R= u_o² Sin 2θ/g here, R = 4km
u_o / g = 2 √4 = 4
When the bullet is fired at an angle of 45 with the horizontal, then the maximum range (R_max) is achieved that is, R_max = u_o/ g
By substituting the values, we get:
R_max = 2 √4 = 4 km
Hence, the bullet will not hit a target 4 km away.

Q7: Calculate the velocity with which an object is shot at an angle of 60° from the ground, and it reaches its maximum height in the 20s. Take g = 10 m/s².
(a) 230.94 m/s
(b) 118 m/s
(c) 126 m/s
(d) 10.55 m/s
Ans: (a)
Here, t = 40, θ = 60°, g = 10 and the time of flight will be in the 40s because the time required for achieving maximum height will be equal to half of the flight time. Since, flight time is t = 2(v sinθ/g), the initial velocity will be 230.94 m/s.

Q8: What is Projectile Motion?
Ans: A particle moves along a curved path under constant acceleration when it is thrown obliquely near the Earth’s surface. This curved path is always directed towards the centre of the Earth. The path of such a particle is called the trajectory of the projectile, and the motion is called projectile motion.

Q9: What is the concept of acceleration in horizontal and vertical projectile motion?
Ans: The only force acting on an object when it is projected in the air with some speed is the acceleration due to gravity (g). There is no acceleration in the horizontal direction because it acts vertically downwards, which means that the particle’s velocity remains constant in the horizontal direction.

Q10: What is the trajectory?
Ans: A trajectory is known as the path followed by a projectile. When an object is thrown into space and upon which the only force acting is the force of gravity, then it is termed as a projectile. This doesn’t mean that other forces are not acting on it, but gravity is the primary force acting on a projectile. Due to this, the effects of other forces are minimised.