1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# A projectile is thrown so as to have a maximum possible horizontal range of 400 m. Taking the point of projection as the origin, the co-ordinates of the point where the velocity of the projectile is minimum is

A
(400,100)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(200,100)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(400,200)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(200,200)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B (200,100)For R to be maximum sin2θ should be maximum ⇒sin2θ=1⇒2θ=90∘⇒θ=45∘ According to question R=u2g=400 m We know that velocity is minimum at the highest point of projectile (i.e. H) by using the relation. Rtanθ=4H 400 tan45∘=4H ⇒H=100 ∴ Coordinates of the point where velocity is minimum is (200,100)

Suggest Corrections
58
Join BYJU'S Learning Program
Related Videos
Solving n Dimensional Problems
PHYSICS
Watch in App
Join BYJU'S Learning Program