Projectile Motion(12th Grade > Physics ) Questions and answers for exam Preparation

Question 1. Four bodies P, Q, R and S are projected with equal velocities having angles of projection

15^{\circ}, 30^{\circ}, 45^{\circ} a n d 60^{\circ}

with the horizontal respectively. The body having shortest range is

P
Q
R
S

Explanation:-

Answer: Option A. -> P
:
A
Range of projectile will be minimum for that angle which is farthest from

45^{\circ}

.

Question 2. Two seconds after projection a projectile is travelling in a direction inclined at

30^{\circ}

to the horizontal after one more sec, it is travelling horizontally, the magnitude and direction of its velocity are

2√20m/sec,60∘
20√3m/sec,60∘
6√40m/sec,30∘
40√6m/sec,30∘

Explanation:-

Answer: Option B. -> 20√3m/sec,60∘
:
B
Let in 2 sec body reaches upto point A and after one more sec upto point B.
Total time of ascent for a body is given 3 sec i.e.

t = \frac{u s i n θ}{g} = 3

∴ u s i n θ = 10 \times 3 = 30 . . . . . . . . . . . . . (i)

Horizontal component of velocity remains always constant

u c o s θ = v c o s 30^{\circ} . . . . . . . . . . (i i)

For vertical upward motion between point O and A

v s i n 30^{\circ} = u s i n θ - g \times 2 [U s i n g v = u - g t v s i n 30^{\circ} = 30 - 20 [A s u s i n θ = 30]

Substituting this value in equation (ii) u coa

θ = 20 c o s 30^{\circ} = 10 \sqrt{3} . . . . . . (i i i)

From equation (i) and (iii)

u = 20 \sqrt{3} a n d θ = 60^{\circ}

Two Seconds After Projection A Projectile Is Travelling In A...

Question 3. The speed of a projectile at the highest point becomes

\frac{1}{\sqrt{2}}

times its initial speed. The horizontal range of the projectile will be

u2g
u22g
u23g
u24g

Explanation:-

Answer: Option A. -> u2g
:
A
Velocity at the highest point is given by

u c o s θ = \frac{u}{\sqrt{2}} (g i v e n) ∴ θ = 45^{\circ}

Horizontal range

R = \frac{u^{2} s i n 2 θ}{g} = \frac{u^{2} s i n (2 \times 45^{\circ})}{g} = \frac{u^{2}}{g}

Question 4. Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths accoding to initial horizontal velocity component, highest first

1, 2, 3, 4
2, 3, 4, 1
3, 4, 1, 2
4, 3, 2, 1

Explanation:-

Answer: Option D. -> 4, 3, 2, 1
:
D
Range

\propto

horizontal component of velocity. Graph 4 shows maximum range, so football possess maximum horizontal velocity in this case.

Question 5. Distance between a frog and an insect on a horizontal plane is

9 m

. Frog can jump with a maximum speed of

\sqrt{10} m / s

. Minimum number of jumps required by the frog to catch the insect is (Take

g = 10 m / s^{2}

)

7
8
9
4

Explanation:-

Answer: Option C. -> 9
:
C
Range is maximum for

45^{\circ}

projection
The distance covered in one hop will be, R =

\frac{u^{2}}{g}

=1 m
Hence, it would take 9 jumps

Question 6. A boy throws a ball with a velocity u at an angle θ with the horizontal. At the same instant he starts running with uniform velocity to catch the ball before it hits the ground. To achieve this, he should run with a velocity of :

u cos θ
u sin θ
u tan θ
√u2tanθ

Explanation:-

Answer: Option A. -> u cos θ
:
A
The horizontal component of velocity of projection and the speed of the boy should be the same.

Question 7. Two seconds after projection a projectile is travelling in a direction inclined at

30^{\circ}

to the horizontal after one more sec, it is travelling horizontally, the magnitude and direction of its velocity are

2√20m/sec,60∘
20√3m/sec,60∘
6√40m/sec,30∘
40√6m/sec,30∘

Explanation:-

Answer: Option B. -> 20√3m/sec,60∘
:
B
Let in 2 sec body reaches upto point A and after one more sec upto point B.
Total time of ascent for a body is given 3 sec i.e.

t = \frac{u s i n θ}{g} = 3

∴ u s i n θ = 10 \times 3 = 30 . . . . . . . . . . . . . (i)

Horizontal component of velocity remains always constant

u c o s θ = v c o s 30^{\circ} . . . . . . . . . . (i i)

For vertical upward motion between point O and A

v s i n 30^{\circ} = u s i n θ - g \times 2 [U s i n g v = u - g t v s i n 30^{\circ} = 30 - 20 [A s u s i n θ = 30]

Substituting this value in equation (ii) u coa

θ = 20 c o s 30^{\circ} = 10 \sqrt{3} . . . . . . (i i i)

From equation (i) and (iii)

u = 20 \sqrt{3} a n d θ = 60^{\circ}

Question 8. A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of

30^{\circ}

with the horizontal. The change in momentum (in magnitude) of the body is

24.5 N–s
49.0 N–s
98.0 N–s
50.0 N–s

Explanation:-

Answer: Option B. -> 49.0 N–s
:
B
Change in momentum between complete projectile motion =

2 m u s i n θ = 2 \times 0.5 \times 98 \times s i n 30^{\circ} = 49 N - s

Question 9. A particle (A) is dropped from a height and another particle (B) is thrown in horizontal direction with speed of 5 m/sec from the same height, bith occuring at the same time. The correct statement is

Both particles will reach at ground simultaneously
Both particles will reach at ground with same speed
Particle (A) will reach at ground first with respect to particle (B)
Particle (B) will reach at ground first with respect to particle (A)

Explanation:-

Answer: Option A. -> Both particles will reach at ground simultaneously
:
A
For both cases t =

\sqrt{\frac{2 h}{g}}

=constant.
Because vertical downward component of velocity will be zero for both the particles.

Question 10. A particle is projected at an angle of

37^{\circ}

with an inclined plane as shown in figure. Calculate:
(i) Time of flight of particle.
(ii) Distance traveled by particle (AB) along the inclined plane

A Particle Is Projected At An Angle Of 37∘ With An Incline...

Time of flight = 125sec Distance along incline = 125(8−6√3)m
Time of flight = 2 sec Distance along incline = 16 m
Time of flight = 3 sec Distance along incline = 6 m
Time of flight = 125sec Distance along incline = (8−6√3)m

Explanation:-

Answer: Option A. -> Time of flight = 125sec Distance along incline = 125(8−6√3)m
:
A
(i) To find out time of flight here, we can analyze the motion in y-direction; we can use the formula

y = u_{y} t + \frac{1}{2} a_{y} t^{2}

. By analyzing motion in y-direction , the displacement of the particle in y-direction during motion is zero.

Now

u_{y} = u s i n α = u . s i n 37^{\circ} = \frac{3}{5} \times 10 = 6 \frac{m}{s}

a_{y} = - g c o s θ = - g c o s 60^{\circ} = - 10 \times \frac{1}{2} = - 5 \frac{m}{s^{2}}

So,

y = u_{y} t + \frac{1}{2} a_{y} t^{2} \Rightarrow 0 = 6 t - \frac{5}{2} t^{2} \Rightarrow t = \frac{12}{5} s

(ii) To find out the distance traveled along AB, we have to analyze the motion in x-direction. So we have to use the formula

x = u_{x} t + \frac{1}{2} a_{x} t^{2}

Here

u_{x} = u c o s α = 10 c o s 37^{\circ} = 10 \times \frac{4}{5} = 8 \frac{m}{s} a_{x} = - g s i n θ = - 10 s i n 60^{\circ} = - 10 \times \frac{\sqrt{3}}{2} = - 5 \sqrt{3} \frac{m}{s^{2}}

And

t = \frac{12}{5} s, x = 8 \times \frac{12}{5} - \frac{1}{2} .5 \sqrt{3} {(\frac{12}{5})}^{2} = \frac{96}{5} - \frac{5 \sqrt{3}}{2} \times \frac{144}{25} = \frac{96}{5} - \frac{72 \sqrt{3}}{5} = \frac{12}{5} (8 - 6 \sqrt{3}) m