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All questions of Projectiles for EmSAT Achieve Exam
How many directions are possible for the same horizontal range?- a)4
- b)3
- c)2
- d)1
Correct answer is option 'C'. Can you explain this answer?
How many directions are possible for the same horizontal range?
a)
4
b)
3
c)
2
d)
1
|
Mira Joshi answered |
We know that the horizontal range for any projectile motion let say R = 2u2.sin 2a /g
Where u is initial speed, and a is the angle at which the particle is thrown, which is responsible for direction. So in between the possible range of a that is 0 - 90, there are maximum two equal values of sin 2a, thus the maximum number of directions for the same or equal range are 2.
Where u is initial speed, and a is the angle at which the particle is thrown, which is responsible for direction. So in between the possible range of a that is 0 - 90, there are maximum two equal values of sin 2a, thus the maximum number of directions for the same or equal range are 2.
If an object is dropped through the window of a fast running train. Then- a)the object moves straight horizontally.
- b)to a man standing near the track, path of the object will be a part of parabola.
- c)the object falls down vertically.
- d)the object will follow an elliptical path
Correct answer is option 'B'. Can you explain this answer?
If an object is dropped through the window of a fast running train. Then
a)
the object moves straight horizontally.
b)
to a man standing near the track, path of the object will be a part of parabola.
c)
the object falls down vertically.
d)
the object will follow an elliptical path
|
Suresh Reddy answered |
When an object is still held it is the part of a system in which the train is moving, once it is left it still has the same velocity as the train for a person from ground. Thus as the velocity is horizontal and acceleration is vertical, it would follow a parabolic trajectory.
The quantity which remains unchanged during the flight of an oblique projectile is- a)Horizontal component of acceleration
- b)Vertical component of acceleration
- c)Vertical component of velocity
- d)Horizontal component of velocity
Correct answer is option 'D'. Can you explain this answer?
The quantity which remains unchanged during the flight of an oblique projectile is
a)
Horizontal component of acceleration
b)
Vertical component of acceleration
c)
Vertical component of velocity
d)
Horizontal component of velocity
|
Sarthak Rajendra Bande answered |
As there is no horizontal acceleration is present in oblique projection horizontal component of velocity is constant
Which statement is true for a ball thrown at 20 degrees with the horizontal, when it is at the highest point in its trajectory?- a)Its velocity is zero but its acceleration is nonzero
- b)Its velocity is nonzero but its acceleration is zero
- c)Its velocity and acceleration are both zero
- d)Its velocity is perpendicular to its acceleration
Correct answer is option 'D'. Can you explain this answer?
Which statement is true for a ball thrown at 20 degrees with the horizontal, when it is at the highest point in its trajectory?
a)
Its velocity is zero but its acceleration is nonzero
b)
Its velocity is nonzero but its acceleration is zero
c)
Its velocity and acceleration are both zero
d)
Its velocity is perpendicular to its acceleration
|
|
Suresh Reddy answered |
At the highest point in its trajectory, the acceleration (g) is acting in downward direction, whereas velocity of the projectile is only in the X (horizontal) direction.
A body is thrown with a velocity of 10m/s at an angle of 60 degrees with the horizontal. Its velocity at the highest point is: - a)5 m/s
- b)8.7 m/s
- c)10 mis
- d)0
Correct answer is option 'A'. Can you explain this answer?
A body is thrown with a velocity of 10m/s at an angle of 60 degrees with the horizontal. Its velocity at the highest point is:
a)
5 m/s
b)
8.7 m/s
c)
10 mis
d)
0
|
|
Mira Joshi answered |
At the highest point, its vertical velocity will become zero. Hence the only left velocity is horizontal which remains unchanged and is equal to 10 cos 60 = 5m/s
A battle ship simultaneously fires two shells at enemy ships. Both are fired with the same speed but with different directions as shown. If the shells follow the parabolic trajectories shown, which ship gets hit first?
- a)they get hit at the same time.
- b)ship μ2
- c)need more information about the trajectories.
- d)ship μ1
Correct answer is 'B'. Can you explain this answer?
A battle ship simultaneously fires two shells at enemy ships. Both are fired with the same speed but with different directions as shown. If the shells follow the parabolic trajectories shown, which ship gets hit first?
a)
they get hit at the same time.
b)
ship μ2
c)
need more information about the trajectories.
d)
ship μ1
|
Krishna Iyer answered |
Ship 2 gets hit first. The time of flight only depends on the y-component of motion, not the x-component. The higher you throw something up in the air, the more time it spends in the air. It can also be shown from equation-
y=v0y t – ½ g t2=0
v0y for projectile μ1 is greater than for μ2.
When a projectile is thrown up at an angle θ to the ground, the time taken by it to rise and to fall are related as- a)time of rise can be less than or equal to time of fall
- b)they are equal
- c)time of rise < time of fall
- d)time of rise > time of fall
Correct answer is option 'B'. Can you explain this answer?
When a projectile is thrown up at an angle θ to the ground, the time taken by it to rise and to fall are related as
a)
time of rise can be less than or equal to time of fall
b)
they are equal
c)
time of rise < time of fall
d)
time of rise > time of fall
|
Gaurav Kumar answered |
The time of fall or rise for a projectile depends upon its vertical component of initial velocity and the vertical acceleration. As the acceleration does not change and the motion of path is identical for the rise phase and the fall phase, we can easily deduce that time taken is also the same.
At the point of maximum height, the acceleration is:- a)minimum
- b)g
- c)maximum
- d)zero
Correct answer is option 'B'. Can you explain this answer?
At the point of maximum height, the acceleration is:
a)
minimum
b)
g
c)
maximum
d)
zero
|
|
Suresh Iyer answered |
At a point of maximum height, the derivative of displacement i.e. velocity is zero but as the gravitational acceleration is equal at all near points to the surface of earth, acceleration at maximum height is still equal to g.
At what angle should a projectile with initial velocity ‘v’ be thrown, so that it achieves its maximum range?- a)450
- b)900
- c)600
- d)300
Correct answer is option 'A'. Can you explain this answer?
At what angle should a projectile with initial velocity ‘v’ be thrown, so that it achieves its maximum range?
a)
450
b)
900
c)
600
d)
300
|
Neha Joshi answered |
Sine of an angle has maximum value 1 when the angle is 90 degree Rmaxis obtained when 2ø = 90 degreeorø= 45 degree
Which one of the following is not an example of projectile?- a)A bullet fired from a gun.
- b)A kicked football.
- c)Taking off of an aircraft.
- d)A javelin thrown by an athlete.
Correct answer is option 'C'. Can you explain this answer?
Which one of the following is not an example of projectile?
a)
A bullet fired from a gun.
b)
A kicked football.
c)
Taking off of an aircraft.
d)
A javelin thrown by an athlete.
|
Top Rankers answered |
A projectile is any object thrown into space upon which the only acting force is gravity. A bullet fired from a gun, a kicked football, and a javelin thrown by an athlete are examples of projectiles once they are in motion and only gravity acts on them. Taking off of an aircraft is not a projectile because it involves continuous external force from the engines.
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