# Intro to Projectile on an Incline

## Trending Questions

**Q.**

A body is projected at$t=0$with a velocity $10m{s}^{\u20131}$ at an angle of $60\xb0$ with the horizontal. The radius of curvature of its trajectory at $t=1s$ is $R$. Neglecting air resistance and taking acceleration due to gravity$g=10m{s}^{\u20132}$, the value of $R$ is :

$2.5m$

$10.3m$

$2.8m$

$5.1m$

**Q.**In figure the angle of inclination of the inclined plane is 30∘. Find the horizontal velocity V0 so that the particle hits the inclined plane perpendicularly.

- V0=√gH5
- V0=√gH7
- V0=√2gH5
- V0=√2gH7

**Q.**

A particle is projected from a point on the surface of smooth inclined plane (see figure). Simultaneously another particle Q is released on the smooth inclined plane from the same position. P and Q collide on the inclined plane after t= 4 second. The speed of projection of P is

**Q.**A particle is projected at an angle α with the horizontal from the foot of a plane, whose inclination to the horizontal is β. What is the value of tan(α−β) if it strikes the plane at right angle?

- cotβ2
- cosec β
- tanβ2
- sinβ2

**Q.**A particle is projected from a point at the foot of a fixed plane inclined at angle 45∘ to the horizontal, in the vertical plane containing the line of greatest slope through the point. If θ(>45∘) is the inclination to the horizontal of the initial direction of projection, for what values of tan θ will the particle strike the plane horizontally?

**Q.**A particle is projected up an inclined plane of inclination β at an elevation α to the horizon. Which of the following are true ?

- If the particle strikes the plane at right angles, then tanα=cotβ+2tanβ
- If the particle strikes the plane horizontally, then tanα=2tanβ
- If the particle strikes the plane at right angles, then tanβ=cotβ+2tanα
- If the particle strikes the plane horizontally, then tanβ=2tanα

**Q.**

A body thrown up along a frictionless inclined plane of angle of inclination 30∘ covers a distance of 40 m along the plane. If the body id projected with the same speed at an angle of 30∘ with the ground, it will have a range of

(Take g=10 ms−2)

20 m

40 m

20√2 m

20√3 m

**Q.**A projectile is fired horizontally from an inclined plane (of inclination 30∘ with the horizontal) with speed 50 m/s. If g=10 m/s2, the range measured along the incline is

- 500 m
- 10003 m
- 200√2 m
- 100 m

**Q.**The horizontal range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be:

- 60 m
- 71 m
- 100 m
- 141 m

**Q.**A 20 kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole, as shown in figure. A cable at an angle of 30∘ with the beam helps to support the light. The horizontal force exerted on the beam by the pole is

**Q.**A particle is projected with certain velocity at an angle α above the horizontal from the foot of an inclined plane having inclination of 30∘. If the particle strikes the plane normally then α is:

- α=30∘+tan−12√3
- 45∘
- 60∘
- α=30∘+tan−1√32

**Q.**A particle is projected at an angle α with the horizontal from the foot of a plane, whose inclination to the horizontal is β. Find the velocity with which the particle strikes perpendicular to the inclined plane.

- ucos(α−β)
- usin(α−β)
- usin(α2)
- ucos(β2)

**Q.**An inclined plane is making an angle β with horizontal. A projectile is projected from the bottom of the plane with a speed u at an angle α with horizontal then its range R when it strikes the inclined plane is :

- 2u2sin(α−β)cosαgcos2β
- 2u2sin(α+β)cosαgcos2β
- u2sin(α−β)cosαgcos2β
- u2sin(α+β)cosαgcos2β

**Q.**In figure the angle of inclination of the inclined plane is 30∘. Find the horizontal velocity V0 so that the particle hits the inclined plane perpendicularly.

- V0=√2gH5
- V0=√2gH7
- V0=√gH5
- V0=√gH7

**Q.**A particle is projected from horizontal making an angle 60∘ with initial velocity 40 m/s. Find the time taken to the particle to make angle 45∘ from horizontal.

**Q.**Two projectiles A and B are projected with velocities √2v and v respectively. They have the same range. If A is thrown at angle of 15o with the horizontal, then what is the angle of projection of B?

**Q.**The range of a projectile fired at an angle of 15 is 50m. If it is fired with the same speed at an angle of 45, its range will be

- 25m
- 37m
- 50m
- 100m

**Q.**A particle is projected with certain velocity at an angle α above the horizontal from the foot of an inclied plane of inclination 300 . If the particle strikes the plane normally then α is

- 300+tan−1(√32)
- 600
- 300+tan−1(2√3)
- 450

**Q.**

A body is projected up a smooth inclined plane (length = 20√2m ) with velocity u from the point M as shown in the figure. The angle of inclination is 45∘ and the top is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v

40 ms−1

40 √2 ms−1

20 ms−1

20 √2 ms−1

**Q.**A particle is projected up an inclined plane of inclination β at an elevation a to the horizon. Show that

a. tanα+cotβ+2tanβ, if the particle strikes the plane at right angles

b. tanα=2tanβ if the particle strikes the plane horizontally.

**Q.**A particle projected from ground moves at 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(√2)
- tan−1(4)

**Q.**A particle of mass m moves according to the equation F=−amr where a is a positive constant, r is radius vector. r=r0^i and v=v0^j at t=0. Describe the trajectory.

- (xr0)2+a(yv0)2=1
- (xr0)2+a(yv0)2=0
- (xr0)2+(yv0)2=1α
- noneofthese

**Q.**In a stationary wave represented by y = asin(ωt)cos(kx) amplitude of the component progressive wave is

- a2
- a
- 2a
- None of these

**Q.**On a horizontal smooth surface a disc is placed at rest. Another disc of same mass is coming with impact parameter equal to its own radius. First disc is of radius r. What should be the radius of coming disc so that after collision first disc moves at an angle 45o to the direction of motion of incoming disc?

- 2r
- r(√2−1)
- r√2
- r(√2−1)

**Q.**It was calculated that a shell when fired from a gun with a certain velocity and at an angle of elevation 5π36 rad should strike a given target in the same horizontal plane. In actual practice, it was found that a hill just prevented the trajectory. At what angle of elevation should the gun be fired to hit the target.

- 5π36rad
- 11π36rad
- 7π36rad
- 13π36rad

**Q.**A cannon is to fire up to 500 m horizontally. What should be the angle of projection, if the shells are fired with a velocity of 100 m/s? (g=10 m/s2)

**Q.**It was calculated that a shell when fired from a gun with certain velocity and at an angle of elevation 5π36 radian should strike a given target. In actual practice, it was found that a hill just intervened the trajectory. The angle of elevation at which the gun should have been fired in order to hit the target is:

- 5π36 radian
- 13π36 radian
- 11π36 radian
- 7π36 radian

**Q.**A body is projected such that its K.E. at the top is (3/4)th of its initial K.E. What is the angle of projectile with the vertical?

**Q.**A particle is projected from the ground at some angle to the horizontal. (Assuming point of projection to be the origin and the horizontal and vertical directions to be the x & y-axis) the particle passes through the points (3, 4) and (4, 3) during its motion then the range of the particle would be:

- 377m
- 36m
- 16m
- 4m

**Q.**

A particle is projected at an angle of 37∘ with an inclined plane. The inclined plane is at an angle of 60∘ with the horizontal. Find

I. Time of flight of particle.

II. Distance traveled by particle (AB) along the inclined plane

I. Time of flight =1.2s

II. Distance traveled =5.7mI. Time of flight =2.4s

II. Distance traveled =5.7mI. Time of flight =4.8s

II. Distance traveled =11.4mI. Time of flight =6.4s

II. Distance traveled =11.4m