Kinematics 5 Question 2

4. A plane is inclined at an angle α=30 with respect to the horizontal. A particle is projected with a speed u=2ms1, from the base of the plane, making an angle θ=15 with respect to the plane as shown in the figure. The distance from the base, at which the particle hits the plane is close to

[Take, g=10ms2 ]

(2019 Main, 10 April I)

(a) 26cm

(b) 20cm

(c) 18cm

(d) 14cm

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Answer:

Correct Answer: 4. (b)

Solution:

  1. When a projectile is projected at an angle θ with an inclined plane making angle α with the horizontal, then

Components of u along and perpendicular to plane are ux=ucosθ and uy=usinθ

We can also resolve acceleration due to gravity into its components along and perpendicular to plane as shown below

So, we can now apply formula for range, i.e. net horizontal displacement of the particle as

R=uxT+12axT2

where, T =time of flight.

Using formula for time of flight, we have

T=2uyay=2usinθgcosα

From Eqs. (i) and (ii), we have

Range up the inclined plane is

R=uxT+12axT2=ucosθ2usinθgcosα12gsinα2usinθgcosα

Here, u=2ms1,g=10ms2,θ=15,α=30

So, T=2usinθgcosα=2×2sin1510×cos30

=2×2×0.258×210×1.732=0.1191

Now,

R=2×cos15×0.119112×10sin30(0.1191)2=2×0.965×0.119152(0.1191)2=0.2290.0354=0.1936m0.20m=20cm



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