at what height id the velocity vector of a projectile motion perpendicular to its initial velocity vector and at what time?????

Lavkesh Singh, 13 years ago

Grade:

1 Answers

SAGAR SINGH - IIT DELHI

878 Points

13 years ago

Dear student,

The projectile motion emphasizes one important aspect of constant acceleration that even constant acceleration, which is essentially unidirectional, is capable to produce two dimensional motion. The basic reason is that force and initial velocity of the object are not along the same direction. The linear motion of the projected object is continuously worked upon by the gravity, which results in the change of both magnitude and direction of the velocity. A change in direction of the velocity ensures that motion is not one dimensional.

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

The velocity in the vertical direction is given by :

v
y

=

u
y

-
g
t

v
y

=

u
y

-
g
t

(2)

An inspection of equation - 2 reveals that this equation can be used to determine velocity in vertical direction at a given time “t” or to determine time of flight “t”, if final vertical velocity is given. This assumes importance as we shall see that final vertical velocity at the maximum height becomes zero.

Figure 5: Vertical component of velocity during motion Projectile motion
Projectile motion (pm5.gif)

The equation for velocity further reveals that the magnitude of velocity is reduced by an amount “gt” after a time interval of “t” during upward motion. The projectile is decelerated in this part of motion (velocity and acceleration are in opposite direction). The reduction in the magnitude of velocity with time means that it becomes zero corresponding to a particular value of “t”. The vertical elevation corresponding to the position, when projectile stops, is maximum height that projectile attains. For this situation (


           v
           y


           v
           y

= 0), the time of flight “t” is obtained as :

v
y

=

u
y

-
g
t

⇒
0
=

u
y

-
g
t

⇒
t
=

u
y

g

v
y

=

u
y

-
g
t

⇒
0
=

u
y

-
g
t

⇒
t
=

u
y

g

(3)

Immediately thereafter, projectile is accelerated in vertically downward direction with increasing speed. In order to appreciate variation of speed and velocity during projectile motion, we calculate the values of a projectile for successive seconds, which is projected with an initial velocity of 60 m/s making an angle of


           30
           0


           30
           0

with the horizontal. Here, vertical component of velocity is 60 sin


           30
           0


           30
           0

" i.e. 30 m/s.