
A projectile is thrown with velocity v making an angle $\theta $ with the horizontal. It just crosses the top of two poles, each of height h, after 1 second and 3 second respectively. The time of flight of the projectile is
(A) 1s
(B) 3s
(C) 4s
(D) 7.8s
Answer
581.7k+ views
- Hint: Projectile motion is a form of motion experienced by an object or particle that is projected near the Earth’s surface and moves along a curved path under the action of gravity only. This curved path was shown by Galileo to be a parabola, but may also be a line in the special case when it is thrown vertically upwards. The study of such motions is called ballistics and such trajectory is called a ballistic trajectory.
Complete step-by-step answer:
Now, form the question-
A projectile is thrown with velocity v making an angle $\theta $ with the horizontal.
Then vertical component of velocity is, ${u_y} = v\sin \theta $
Now applying formula
$s = ut + \dfrac{1}{2}a{t^2}$
For vertical direction,
s = y, $u = {u_y}$, $a = {a_y} = - g$
Substituting these values in above equation,
$y = v\sin \theta - \dfrac{1}{2}g{t^2}$
If we put t = 1sec then y = h
$h = v\sin \theta - \dfrac{1}{2}g$………(1)
Again, if we put t = 3sec then y = h
So, $h = 3v\sin \theta - \dfrac{9}{2}g$………(2)
Equating equations (1) and (2), we get
$2v\sin \theta = 4g$
Now, time of flight (T) = $\dfrac{{2v\sin \theta }}{g}$
$ \Rightarrow T = \dfrac{{4g}}{g} = 4\sec $ [since, 2vsin$\theta $ = 4g]
Hence the correct option is $C.$
Note: The time of flight of a projectile motion is the time from when the object is projected to the time it reaches the surface.
The key components that we need to remember in order to solve projectile motion problems are:
Initial launch angle, $\theta $
Initial velocity, u
Time of flight, T
Acceleration, a
Horizontal velocity, ${u_x}$
Vertical velocity, ${u_y}$
Displacement, d
Maximum height, H
Range, R
Projectile motion is when an object moves in a bilaterally symmetrical, parabolic path. The path that the object follows is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning, after which the only influence on the trajectory is that of gravity. Objects that are projected from and land on the same horizontal surface will have a vertically symmetrical path. The time of flight depends on the initial velocity of the projectile and the angle of projection.
Complete step-by-step answer:
Now, form the question-
A projectile is thrown with velocity v making an angle $\theta $ with the horizontal.
Then vertical component of velocity is, ${u_y} = v\sin \theta $
Now applying formula
$s = ut + \dfrac{1}{2}a{t^2}$
For vertical direction,
s = y, $u = {u_y}$, $a = {a_y} = - g$
Substituting these values in above equation,
$y = v\sin \theta - \dfrac{1}{2}g{t^2}$
If we put t = 1sec then y = h
$h = v\sin \theta - \dfrac{1}{2}g$………(1)
Again, if we put t = 3sec then y = h
So, $h = 3v\sin \theta - \dfrac{9}{2}g$………(2)
Equating equations (1) and (2), we get
$2v\sin \theta = 4g$
Now, time of flight (T) = $\dfrac{{2v\sin \theta }}{g}$
$ \Rightarrow T = \dfrac{{4g}}{g} = 4\sec $ [since, 2vsin$\theta $ = 4g]
Hence the correct option is $C.$
Note: The time of flight of a projectile motion is the time from when the object is projected to the time it reaches the surface.
The key components that we need to remember in order to solve projectile motion problems are:
Initial launch angle, $\theta $
Initial velocity, u
Time of flight, T
Acceleration, a
Horizontal velocity, ${u_x}$
Vertical velocity, ${u_y}$
Displacement, d
Maximum height, H
Range, R
Projectile motion is when an object moves in a bilaterally symmetrical, parabolic path. The path that the object follows is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning, after which the only influence on the trajectory is that of gravity. Objects that are projected from and land on the same horizontal surface will have a vertically symmetrical path. The time of flight depends on the initial velocity of the projectile and the angle of projection.
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