Physics Multiple Choice Questions on “Simple Projectile Motion”.
1. A body of mass m, projected at an angle of θ from the ground with an initial velocity of v, acceleration due to gravity is g, what is the maximum horizontal range covered?
a) R = v2 (sin 2θ)/g
b) R = v2 (sin θ)/2g
c) R = v2 (sin 2θ)/2g
d) R = v2 (sin θ)/g
Answer: a
Clarification: The calculation for range uses the horizontal component of the initial velocity and time. Range = distance x Time taken for the motion. The time can be found by utilising the fact that the vertical velocity becomes zero at the maximum height (i.e. at half of the time period) and using the first equation of motion for this. Hence, on calculating, the range will be obtained as R = v2 (sin 2θ)/g.
2. A body of mass 5 kg, projected at an angle of 60° from the ground with an initial velocity of 25 m/s, acceleration due to gravity is g = 10 m/s2, what is the maximum horizontal range covered?
a) 54.13 m
b) 49 m
c) 49.16 m
d) 60 m
Answer: a
Clarification: The formula for horizontal range is R = v2 (sin 2θ)/g. Here, v = 25, θ = 60°, g = 10. Hence, on solving we will get range as 54.13 m.
3. A body of mass 10 kg, projected at an angle of 30° from the ground with an initial velocity of 5 m/s, acceleration due to gravity is g = 10 m/s2, what is the time of flight?
a) 0.866s
b) 1.86 s
c) 1.96 s
d) 1.862 s
Answer: a
Clarification: The formula for time of flight is t = 2(v sinθ/g). Here, v = 5, θ = 30°, g = 10. Hence, on solving we will get the time of flight as 0.866 s.
4. On calculating which of the following quantities, the mass of the body has an effect in simple projectile motion?
a) Velocity
b) Force
c) Time of flight
d) Range
Answer: b
Clarification: All mentioned quantities except force are kinematic quantities. Force is a kinetic quantity. Force = m x a. Hence, the mass of the body has an effect on force calculation.
5. A ball of mass 100 g, projected at an angle of 30° from the ground with an initial velocity of 11 m/s, acceleration due to gravity is g = 10 m/s2, what is the maximum height attained?
a) 1.5 m
b) 3.0 m
c) 1.0 m
d) 2.0 m
Answer: a
Clarification: The formula for maximum height is h = (v sinθ)2/2g. Here, v = 11, θ = 30°, g = 10. Hence, on solving we will get the maximum height attained as 1.5125 m. The value can be rounded off to 1.5.
6. When do we get maximum range in a simple projectile motion?
a) When θ = 45°
b) When θ = 60°
c) When θ = 90°
d) When θ = 0°
Answer: a
Clarification: The formula for horizontal range is R = v2(sin 2θ)/g. This will be maximum when sin 2θ = 1, which implies that 2θ = 90°, which in turn implies that θ = 45°. Hence the correct answer is when θ = 45°.
7. When do we get maximum height in a simple projectile motion?
a) When θ = 45°
b) When θ = 60°
c) When θ = 90°
d) When θ = 0°
Answer: c
Clarification: The formula for horizontal range is h = (v sinθ)2/2g. This will be maximum when sin θ = 1, which implies that θ = 90°. Hence the correct answer is when θ = 90°.
8. A football is projected at an angle of 45° from the ground with an initial velocity of 10 m/s, take acceleration due to gravity is g = 10 m/s2.What is the time of flight?
a) 1.4142 s
b) 1.5361 s
c) 1.8987 s
d) 1.5651 s
Answer: a
Clarification: The formula for time of flight is t = 2(v sinθ/g). Here, v = 10, θ = 45°, g = 10. Hence, on solving we will get the time of flight as 1.4142 s.
9. A bag of mass 1000 g, projected at an angle of 90° from the ground with an initial velocity of 5 m/s, acceleration due to gravity is g = 10 m/s2, what is the maximum height attained?
a) 1.25 m
b) 3.0 m
c) 1.5 m
d) 2.0 m
Answer: a
Clarification: The formula for maximum height is h = (v sinθ)2/2g. Here, v = 5, θ = 90°, g = 10. Hence, on solving we will get the maximum height attained as 1.25 m. The maximum height can alternatively be found by simply using equations of motion as this bag is thrown vertically upwards.
10. A body of mass 55 kg, projected at an angle of 45° from the ground with an initial velocity of 15 m/s, acceleration due to gravity is g = 10 m/s2, what is the maximum horizontal range covered?
a) 22.5 m
b) 25 m
c) 16 m
d) 15 m
Answer: a
Clarification: The formula for horizontal range is R = v2 (sin 2θ)/g. Here, v = 15, θ = 45°, g = 10. Hence, on solving we will get range as 22.5 m.
11. At what angle of projectile (θ) is the horizontal range minimum?
a) θ = 45°
b) θ = 60°
c) θ = 90°
d) θ = 75°
Answer: c
Clarification: The formula for horizontal range is R = v2 (sin 2θ)/g. When θ = 90°, sin 2θ = sin 180° = 0. Hence, the range covered becomes 0. Since range cannot be negative, 0 is the minimum value it can attain.
12. A body of mass 5 kg, projected at an angle of 45° from the ground covers a horizontal range of 45 m, acceleration due to gravity is g = 10 m/s2, what is the velocity with which it was projected covered?
a) 21.21 m/s
b) 20 m/s
c) 22 m/s
d) 21.1 m/s
Answer: a
Clarification: The formula for horizontal range is R = v2 (sin 2θ)/g. Here, R = 45, θ = 45°, g = 10. Hence, on solving we will get velocity as 21.21 m/s. The result can be cross checked by putting this value of velocity and finding out the range.
13. A big stone of mass 1000 g, projected at an angle of 30° from the ground it covers a maximum vertical distance of 5 m, acceleration due to gravity is g = 10 m/s2, what is the velocity with which it was thrown?
a) 11.55 m/s
b) 11.5 m/s
c) 1.155 m/s
d) 12.0 m/s
Answer: a
Clarification: The maximum vertical distance is the maximum height. The formula for maximum height is h = (v sinθ) 2/2g. Here, h = 5, θ = 30°, g = 10. Hence, on solving we will get the initial velocity as 11.55 m/s. The result can be cross checked by putting this value of velocity and finding out the maximum height.
14. A soccer ball is projected at an angle of 60° from the ground.It attains its maximum height in 10s. Considering acceleration due to gravity as g = 10 m/s2. What is the velocity with which it was projected?
a) 115.5 m/s
b) 117 m/s
c) 120 m/s
d) 11.55 m/s
Answer: a
Clarification: The time for achieving the maximum height is equal to half of the time of flight. Therefore, the time of flight is 20s. The formula for time of flight is t = 2(v sinθ/g). Here, t = 20, θ = 60°, g = 10. Hence, on solving we will get the initial velocity as 115.5m/s.