Question

a particle is projected at an angle of projection 60 degree and initial velocity is 30 m/s find the height at which the velocity becomes 20m/s​

Answer 1

Answer:

Answer

Given : v=20m/s

∴ u

x

=vcos60

o

=20×0.5=10m/s (which remains the constant all the time)

Also u

y

=vsin60

o

=20×0.866=17.32m/s (which decreases with time)

Let the time be t when the final velocity is half of initial velocity i.e V

f

=

2

20

=10m/s

Thus at time t, the y component of the velocity is zero i.e V

y

=0

∴ 0=u

y

−gt

OR t=

g

u

y

=

10

17.32

=1.732 s