DC Pandey Solutions: Projectile Motion - 2

# DC Pandey Solutions: Projectile Motion - 2 | Physics Class 11 - NEET PDF Download

Introductory Exercise 4.2

Ques 1: A particle is projected along an inclined plane as shown in figure. What is the speed of the particle when it collides at point A? (g = 10 m/s2)

Ans:
Sol: Time of flight

Using,  v = u + at
vx = ux = u cos 60°

Ques 2: In the above problem what is the component of its velocity perpendicular to the plane when it strikes at A?
Ans: 5 m/s
Sol: Component of velocity perpendicular to plane

Ques 3: Two particles A and B are projected simultaneously from the two towers of height 10 m and 20 m respectively. Particle A is projected with an initial speed of 10√2 m/s at an angle of 45° with horizontal, while particle B is projected horizontally with speed 10 m/s. If they collide in air, what is the distance d between the towers?

Ans: 20 m
Sol: Let the particle collide at time t.

x1 = (u cos θ) t
and x2 = vt
∴ d = x2 - x1
= (v + u cos θ) t

For vertical motion of particle 1:

i.e.,   …(i)
or
For the vertical motion of particle 2:

i.e.,   …(ii)
Comparing Eqs. (i) and (ii),

⇒ t = 1 s
∴ d = 20 m

Ques 4: Two particles A and B are projected from ground towards each other with speeds 10 m/s and 5√2 m/s at angles 30° and 45° with horizontal from two points separated by a distance of 15 m. Will they collide or not?

Ans: No
Sol: u = 10 m/s
v = 5√2 m/s

θ = 30°
φ = 45°
d = 15 m
Let the particles meet (or are in the same vertical time t).
∴ d = (u cos θ) t + (v cos φ) t
⇒ 15 = (10 cos 30° + 5√2 cos 45°) t
or
or
= 1.009 s
Now, let us find time of flight of A and B

= 1 s
As TA < t, particle A will touch ground before the expected time t of collision.
∴ Ans: NO.

Ques 5: A particle is projected from the bottom of an inclined plane of inclination 30°. At what angle α (from the horizontal) should the particle be projected to get the maximum range on the inclined plane.
Ans: 60°
Sol: For range to be maximum

Ques 6: A particle is projected from the bottom of an inclined plane of inclination 30° with velocity of 40 m/s at an angle of 60° with horizontal. Find the speed of the particle when its velocity vector is parallel to the plane. Take g = 10 m/s2.
Ans:
Sol: At point A velocity  of the particle will be parallel to the inclined plane.

∴ φ = β
vx = ux = u cos α
vx = v cos φ = v cos β
or u cos α = v cos β

Ques 7: Two particles A and B are projected simultaneously in the directions shown in figure with velocities vA = 20 m/s and vB = 10 m/s respectively. They collide in air after 1/2 s.
Find:
(a) the angle θ
(b) the distance x.

Ans: (a) 30°

Sol: (a) At time t, vertical displacement of A
= Vertical displacement of B

i.e.,  vA sin θ = vB

∴ θ = 30°
(b) x = (vA cos θ) t

The document DC Pandey Solutions: Projectile Motion - 2 | Physics Class 11 - NEET is a part of the NEET Course Physics Class 11.
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## Physics Class 11

102 videos|411 docs|121 tests

## FAQs on DC Pandey Solutions: Projectile Motion - 2 - Physics Class 11 - NEET

 1. How can I calculate the maximum height reached by a projectile?
Ans. To calculate the maximum height reached by a projectile, you can use the formula: Maximum height = (Initial velocity in the vertical direction)^2 / (2 * acceleration due to gravity)
 2. What is the range of a projectile?
Ans. The range of a projectile is the horizontal distance covered by it before hitting the ground. It can be calculated using the formula: Range = (Initial velocity in the horizontal direction) * (Time of flight)
 3. How does the angle of projection affect the range of a projectile?
Ans. The angle of projection affects the range of a projectile as the range is maximum when the angle is 45 degrees. If the angle is less than 45 degrees, the range decreases, and if it is greater than 45 degrees, the range also decreases.
 4. What are the different equations of motion for projectile motion?
Ans. The equations of motion for projectile motion are: - Vertical motion equation: y = ut * sin(theta) - (1/2) * g * t^2 - Horizontal motion equation: x = ut * cos(theta) * t - Range equation: R = (u^2 * sin(2*theta)) / g - Maximum height equation: H = (u^2 * sin^2(theta)) / 2g
 5. How does air resistance affect projectile motion?
Ans. In most cases, air resistance is negligible in projectile motion calculations. However, if air resistance is significant, it can affect the range and trajectory of the projectile. It can cause the projectile to experience a decrease in velocity and deviation from its ideal path.

## Physics Class 11

102 videos|411 docs|121 tests

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