Question

A ball is thrown with 50m/s at 30^(@) with the horizontal from a building of height 500m.The time (in sec) at which velocity vector of the ball becomes perpendicular to the initial velocity is (g=10m/s^(2)) :-​

Answer 1

Explanation:

During the projectile motion:

Range of particle =

g

u

2

sin2θ

(u= initial velocity of particle, θ= angle of projection)

⇒R=

g

40

2

sin60

o

(u=40m/s and θ=60

o

)

⇒R=

2×10

1600

3

=80

3

m

For downward motion we have: u

y

=40sin30

o

=20m/s

So applying h=ut+

2

1

gt

2

⇒60m=20t+5t

2

⇒t

2

+4t−12=0⇒(t−2)(t+6)=0

⇒t=2s or t=−6s (physically not possible)

So time of descent =2s

Horizontal component of velocity =40cos30

o

=20

3

m/s

So range covered during descent =20

3

(2)=40

3

m

So total horizontal range covered by particle =80

3

+40

3

=120

3

m.