# Projectile Time, Height and Range

## Trending Questions

**Q.**

When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60°with horizontal, it can travel a distance x1 along the plane. But, when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x2 distance. Then x1:x2 will be

- 1:2
- 2:1
- 1:√3
- 1:2√3

**Q.**A particle is projected with a velocity v so that its range on a horizontal plane is twice the greatest height attained. Consider g=10 m/s2 as acceleration due to gravity. Then, its range will be:

- 4v25g
- 4v25g2
- 4v5g2
- 4v35g2

**Q.**A particle is projected at an angle of 60∘ above the horizontal with a speed of 10 m/s. After some time, the direction of its velocity makes an angle of 30∘ above the horizontal. The speed of the particle at this instant is

- 5√3 m/s
- 5√3 m/s
- 5 m/s
- 10√3 m/s

**Q.**

A projectile is fired from the surface of the earth with a velocity of 5 ms−1 and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3 ms−1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet (in ms−2) is

(Given g=9.8 ms−2)

- 3.5
- 5.9
- 16.3
- 110.8

**Q.**The trajectory of a projectile in a vertical plane is given by y=ax−bx2, where a and b are constants and x and y are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are

- b22a, tan−1(b)
- a2b, tan−1(2a)
- a24b, tan−1(a)
- 2a2b, tan−1(a)

**Q.**A metallic rod of mass per unit length 0.5 kg mâˆ’1 is lying horizontally on a smooth inclined plane which makes an angle of 30o with the horizontal. The rod is not allowed to slide down by a passing a current through it when a magnetic field of induction 0.25 T is acting on it in the vertical direction. The current flowing in the rod to keep it stationary is

- 14.76 A

7.14 A

11.32 A

5.98 A

**Q.**The maximum range of a gun on a horizontal terrain is 16 km. If g=10 m/s2, then muzzle velocity of the shell must be

- 160 m/s
- 200√2 m/s
- 400 m/s
- 800 m/s

**Q.**The horizontal range and maximum height attained by a projectile are R and H, respectively. If a constant horizontal acceleration a=g4 is imparted to the projectile due to wind, then its horizontal range and maximum height will be

- (R+H), H2
- (R+H), 2H
- (R+2H), H
- (R+H), H

**Q.**At what angle with the horizontal should a ball be thrown so that the range R is related to the time of flight (T) as R=5T2?(Take g=10 m/s2)

- 60∘
- 30∘
- 45∘
- 90∘

**Q.**A train is moving along a straight line with a constant acceleration. A boy standing inside the train throws a ball in forward direction with a speed of 10 m/s at an angle of 60∘ to the horizontal, with respect to himself. The boy has to move forward by 1.15 m inside the train to catch the ball (at the initial height). Then,

[Take g=10 m/s2]

- Acceleration of train is 5 m/s2
- Acceleration of ball is 10 m/s2
- Acceleration of ball is 5√5 m/s2
- Displacement of train is 8.66 m

**Q.**A ball is projected from the ground at an angled of 45o with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30o with the horizontal surface. The maximum height it reaches after the bounce, (in metres), is

**Q.**The position coordinates of a projectile projected from ground on a certain planet (with no atmosphere) are given by y=(4t−2t2) m and x=(3t) m, where t is in second and point of projection is taken as origin. The angle of projection of projectile with vertical is

- 30∘
- 37∘
- 45∘
- 60∘

**Q.**An object is projected with a velocity of 20 m/s making an angle of 45∘ with horizontal. The equation for the trajectory is h = Ax – Bx2 where h is height, x is horizontal distance, A and B are constants. The ratio A : B is (g = 10 ms−2)

- 1 : 5
- 1 : 40
- 40 : 1
- 5 : 1

**Q.**For a projectile thrown with a speed v, the horizontal range is √3v22g. The vertical range is v28g. The angle which the projectile makes with the horizontal initially is

- 15∘
- 30∘
- 45∘
- 60∘

**Q.**A particle is projected from the ground with velocity u at an angle θ with horizontal. The horizontal range, maximum height and time of flight are R, H and T respectively. They are given by, R=u2sin2θg, H=u2sin2θ2g and T=2usinθg

Now keeping u as fixed, θ is varied from 30∘ to 60∘. Then

- R will first increase then decrease, H will increase and T will decrease
- R will first increase then decrease while H and T both will increase
- R will decrease while H and T will increase
- R will increase while H and T will increase

**Q.**If R and h represent the horizontal range and maximum height respectively of an oblique projectile, then R28h+2h represents

- Maximum horizontal range
- Time of flight
- Maximum verticle range
- Velocity of projectile at highest point.

**Q.**A projectile can have the same range for two angles of projection. If h1 and h2 are maximum heights when the range in the two cases is R, then the relation between R, h1 and h2 is

- R=4√h1h2
- R=2√h1h2
- R=√h1h2
- None of these

**Q.**A particle starts from the origin of co-ordinates at time t=0 and moves in the xy plane with a constant acceleration α in the y – direction. Its equation of motion y=βx2. Its velocity component in the x-direction is

- αβ
- √2αβ
- α2β
- √α2β

**Q.**

A body of mass 1 kg is projected from ground at an angle 30∘ with horizontal on a level ground at a speed 50 m/s. The magnitude of change in momentum of the body during its flight is (g=10 m/s2)

- 50 kg ms−1
- 100 kg ms−1
- 25 kg ms−1
- Zero

**Q.**At a height of 0.4 m from the ground, the velocity of a projectile in vector form is→v=(6^i+2^j) m/s. The angle of projection is

(g=10 m/s2)

- 30∘
- 60∘
- 45∘
- 37∘

**Q.**Ratio of minimum kinetic energies of two projectiles of same mass is 4:1. The ratio of the maximum height attained by them is also 4:1. The ratio of their ranges would be

- 16:1
- 4:1
- 8:1
- 2:1

**Q.**The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

- tan−1(14)
- tan−1(2)
- π4
- tan−1(4)

**Q.**A particle projected from ground moves at an angle 45∘ with horizontal one second after projection and speed is minimum two seconds after the projection. The angle of projection of particle is

[Neglect the effect of air resistance]

- tan−1(3)
- tan−1(2)
- tan−1(4)
- tan−1(√2)

**Q.**A particle is projected at an angle of 30∘ w.r.t horizontal with speed 20 m/s. Take the point of projection as the origin of the coordinate axes and find the velocity vector of the particle after 1 s (in m/s).

- 5√3^i+5^j
- 5^j
- 5^i+10√3^j
- 10√3^i

**Q.**A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected at an angle of 60∘ with the vertical with the same initial speed. At the highest point, the ratio of their potential energies will be

- 3:2
- 2:3
- 2:1
- 4:1

**Q.**A body is projected at an angle of 30∘ with the horizontal and with a speed of 30 ms−1. What is the angle with the horizontal after 1.5 seconds ?

(g=10 ms−2).

- 0∘
- 30∘
- 45∘
- 60∘

**Q.**The velocity at the maximum height of a projectile is √32 times its initial velocity of projection (u). Its range on the horizontal plane is

- √3u22g
- 3u22g
- 3u2g
- u22g

**Q.**The equation of a projectile is y=ax−bx2. Its horizontal range is

- ab
- ba
- a+b
- b−a

**Q.**

A player kicks a football at an angle of 45∘ with a velocity of 20 ms−1. A second player 60 m away along the direction of kick starts running to receive the ball at that instant. Find the speed with which second player should run to reach the ball before it hits the ground (g=10 ms−2)

- 5√2 m/s
- √2 m/s
- 3√2 m/s
- 1 m/s

**Q.**A particle is fired at an angle of 30∘ to the horizontal such that the vertical component of its initial velocity is 80 m/s. Its time of flight is T. Its velocity at t=T4 has a magnitude of nearly

(Take g=10 m/s2)

- 200 m/s
- 300 m/s
- 144 m/s
- 100 m/s