IIT JEE , AIEEE Forum - Physics , Mechanics

jatinpreet

Please answer this question -

A ball is projected from ground with a speed of 10 m/s at an angle 30 degrees with the horizontal at an instant t=0. Find the time elapsed when the velocity vector becomes perpendicular to the initial velocity vector.

Anup

i'm not sure but i think the velocity vector will be perpendicular to the initial vector when it just touches the ground again.....

so time taken =time of flight =2usin

/g

=1second

sorry if its wrong...

jatinpreet

thanks for the reply yaar but i got the correct answer from somewhere . actually when velocity (v) vector is equal to the initial velocity vector (u) then- u.v=0

where u= 10cos30 i + 10 sin30 j
& v= 10cos30 i + (10sin30 - gt)

ac

The approach given by Jatinpreet is conceptually correct:
When final velocity vector is perpendicular to initial velocity vector,
U.V = 0 ---- (1)
If t₁ is the time elapsed until Y component of velocity is zero,
then 0 = 10 Sin30 - gt₁

t₁ = sin30 = 1/2 sec.

If t₂ be the additional time till velocity vector becomes perpendicular to the initial,
V = 10 Cos30 i - gt₂ j ---- (2)
Now, U = 10 Cos30 i + 10 Sin30 j ---- (3)

From (1), (2), and (3) above, we have,
(10 Cos30)² - (10 Sin30)gt₂= 0
Therefore, t₂ = 3/2 sec
Hence total time elapsed = t₁ + t₂ = 2 sec.

But, total time of flight = (2x10xsin30)/g = 1 sec.
Therefore, such a situation will not be reached when the vectors will be perpendicular.

ac

Here's a more straight forward solution for this problem:

Consider the motion of the projectile:

At the middle of the motion, the velocity has only a horizontal component. So the angle between initial velocity and current velocity vectors is 30 degrees. When the ball hits the ground, the angle between the initial velocity and final velocity vectors is 60 degrees. So the angle doesn't exceed 60 degrees during the entire projectile motion. Hence the velocity vectors will never be perpendicular.

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