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

12th Grade > Physics

PROJECTILE MOTION MCQs

Total Questions : 45 | Page 4 of 5 pages

Question 31. If for a given angle of projection, the horizontal range is doubled, the time of flight becomes

4 times
2 times
√2 times
1√2 times

Discuss Question

Answer: Option C. -> √2 times
:
C

R = \frac{u^{2} s i n 2 θ}{g} a n d T = \frac{2 u s i n θ}{g}

∴ R \propto u^{2} a n d T \propto u

(If

θ

and g are constant ).
In the given condition to make range double, velocity must be increased upto

\sqrt{2}

times that of previous value. So automatically time of flight becomes

\sqrt{2}

times.

Question 32. A ball is projected upwards from the top of tower with a velocity

50 m s^{- 1}

making angle

30^{\circ}

with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground

2.33 sec
5.33 sec
7.11 sec
6.33 sec

Discuss Question

Answer: Option C. -> 7.11 sec
:
C
Formula for calculation of time to reach the body on the ground from the tower of height ‘h’ (If it is thrown vertically up with velocity u) is given by

t = \frac{u}{g} [1 + \sqrt{1 + \frac{2 g h}{u^{2}}}]

So we can resolve the given velocity in vertical direction and can apply the above formula.
Initial vertical component of velocity

u s i n θ = 50 s i n 30 = 25 m / s

∴ t = \frac{25}{9.8} [1 + \sqrt{1 + \frac{2 \times 9.8 \times 70}{(25)^{2}}}] = 7.11 s e c

A Ball Is Projected Upwards From The Top Of Tower With A Vel...

Question 33. A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle of

30^{\circ}

with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground

(g = 10 m / s^{2}, s i n 30^{\circ} = \frac{1}{2}, c o s 30^{\circ} = \frac{\sqrt{3}}{2})

8.66 m
5.20 m
4.33 m
2.60 m

Discuss Question

Answer: Option A. -> 8.66 m
:
A
Simply we have to calculate the range of projectile

R = \frac{u^{2} s i n 2 θ}{g} = \frac{(10)^{2} s i n (2 \times 30^{\circ})}{10} R = 5 \sqrt{3} = 8.66 m e t e r

A Boy Playing On The Roof Of A 10m High Building Throws A Ba...

Question 34. A body is projected up a smooth inclined plane (length =

20 \sqrt{2} m

) with velocity u from the point M as shown in the figure. The angle of inclination is

45^{\circ}

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

A Body Is Projected Up A Smooth Inclined Plane (length = 20�...

40 ms−1
40 √2 ms−1
20 ms−1
20 √2 ms−1

Discuss Question

Answer: Option D. -> 20 √2 ms−1
:
D
At point N angle of projection of the body will be

45^{\circ}

. Let velocity of projection at this point is v. If the body just manages to cross the well then
Range = Diameter of well

\frac{v^{2} s i n 2 θ}{g} = 40 [A s θ = 45^{\circ}] v^{2} = 400 \Rightarrow v = 20 m / s

But we have to calculate the velocity (u) of the body at point M.
For motion along the inclined plane (from M to N)
Final velocity (v) = 20 m/s, acceleration (a) =

- g s i n α = - g s i n 45^{\circ}

, distance of inclined plane (s) =

20 \sqrt{2} m

(20)^{2} = u^{2} - 2 \frac{g}{\sqrt{2}} .20 \sqrt{2} [U s i n g v^{2} = u^{2} + 2 a s] u^{2} = 20^{2} + 400 \Rightarrow u = 20 \sqrt{2} m / s .

Question 35. A car is travelling on a highway at a speed of 25 m/s along the x-axis. A passenger in a car throws a ball at an angle

37^{\circ}

with horizontal in a plane perpendicular the motion of the car. The ball is projected with a speed of 10 m/s relative to the car. What may be the initial velocity of the ball in unit vector notation?

A Car Is Travelling On A Highway At A Speed Of 25 M/s Along ...

25 ^i + 8^j + 6^k
10 ^i + 8^j + 6^k
10 ^i + 25^j + 6^k
25 ^i + 6^j + 8^k

Discuss Question

Answer: Option D. -> 25 ^i + 6^j + 8^k
:
D

u_{x}

25^i

as the car is moving with a speed of 25

\frac{m}{s}

along +ve axis, given

θ = 37^{\circ}

∴

\frac{u_{y}}{u_{z}} = t a n 37^{\circ} = \frac{3}{4}

∴

the only possible option is

25^i + 6^j + 8^k

Question 36. A particle is projected from a point O with a velocity u in a direction making an angle

α

upward with the horizontal. After some time at point P it is moving at right angle with its initial direction of projection. The time of flight from O to P is

u sin αg
u cosec αg
u tan αg
u sec αg

Discuss Question

Answer: Option B. -> u cosec αg
:
B
When body projected with initial velocity

⃗ u

by making angle

α

with the horizontal. Then after time t, (at point P) it’s direction is perpendicular to

⃗ u

Magnitude of velocity at point P is given by v =

u c o t α

(from sample problem no. 9)
For vertical motion : Initial velocity (at point O)

= u s i n α

Final velocity (at point P)

= - v c o s α = - u c o t α c o s α

Time of flight (from point O to P) = t
Applying first equation of motion

v = u - g t

- u c o t α c o s α = u s i n α - g t ∴ t = \frac{u s i n α + u c o t α c o s α}{g} = \frac{u}{g s i n α} [s i n^{2} α + c o s^{2} α] = \frac{u c o s e c α}{g}

A Particle Is Projected From A Point O With A Velocity U In ...

Question 37. A ball is projected upwards from the top of tower with a velocity

50 m s^{- 1}

making angle

30^{\circ}

with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground

2.33 sec
5.33 sec
7.11 sec
6.33 sec

Discuss Question

Answer: Option C. -> 7.11 sec
:
C
Formula for calculation of time to reach the body on the ground from the tower of height ‘h’ (If it is thrown vertically up with velocity u) is given by

t = \frac{u}{g} [1 + \sqrt{1 + \frac{2 g h}{u^{2}}}]

So we can resolve the given velocity in vertical direction and can apply the above formula.
Initial vertical component of velocity

u s i n θ = 50 s i n 30 = 25 m / s

∴ t = \frac{25}{9.8} [1 + \sqrt{1 + \frac{2 \times 9.8 \times 70}{(25)^{2}}}] = 7.11 s e c

Question 38. Two projectiles are thrown simultaneously in the same plane from the same point. If their velocities are

v_{1}

and

v_{2}

at angles

θ_{1}

and

θ_{2}

respectively from the horizontal, then answer the following questions: If

v_{1}

v_{2}

and

θ_{1} > θ_{2}

, then choose the incorrect statement

Particle 2 moves under the particle 1
The slope of the trajectory of particle 2 with respect to 1 is always positive
Both the particles will have the same range if θ1 > 45∘ and θ2
none of these

Discuss Question

Answer: Option B. -> The slope of the trajectory of particle 2 with respect to 1 is always positive
:
B

Two Projectiles Are Thrown Simultaneously In The Same Plane ...

Initial velocity of 2 w.r.t. 1:

V_{2 / 1} = (v_{2} c o s θ_{2} - v_{1} c o s θ_{1})^i + (v_{2} s i n θ_{2} - v_{1} s i n θ_{1})^j

Its

^i

part is positive and

^j

part is negative.
Trajectory of particle 2 will be a straight line OA w.r.t 1 as shown. Its slope is negative.
Line 1 - 2 will be parallel to OA at any time.

Question 39. A particle is projected vertically upwards from

O

with velocity

v

and a second particle is projected at the same instant from

P

(at a height h above

O

) with velocity

v

at an angle of projection

θ

. The time when the distance between them is minimum is

h2v sinθ
h2v cosθ
hv
h2v

Discuss Question

Answer: Option D. -> h2v
:
D
Relative acceleration between the two particles is zero. The distance between then at time

t

A Particle Is Projected Vertically Upwards From O With Veloc...

s

\sqrt{(h - (v - v s i n θ t))^{2} + (v c o s θ t)^{2}}

s^{2}

(h - (v - v s i n θ t))^{2}

(v c o s θ t)^{2} s

is minimum when
or

\frac{d}{d t} (s^{2})

0

2 (h - (v - v s i n θ) t) (v s i n θ - v) + 2 v^{2} c o s^{2} θ t

0

t

\frac{h}{2 v}

Question 40. Given the speed-time graph of a golf ball shot from ground at t = 0.
What will be the time of flight?

1s
0.8s
0.6s
1.2s

Discuss Question

Answer: Option B. -> 0.8s
:
B
at t = 0, speed is 5

\frac{m}{s}

, this means, u = 5

\frac{m}{s}

. Also minimum speed is 3

\frac{m}{s}

. For a projectile speed is minimum and equal to its horizontal component at the highest point of its trajectory, as the vertical component is zero at this point. Let