Question

Q24.(i)At what points on the path of an oblique projectile is its velocity(a) maximum and (b) minimum .What are the values of these. (iii)Prove that the maximum horizontal range is four times the maximum height attained by a projectile which is fired along the required oblique direction.​

Answer 1

At the time t=0, the projectile has the maximum speed vertically. At the maximum height the projectile has minimum θ vertical speed.

ϕ= angle of projection

So minimum speed =ucosϕ=10m/s

Maximum speed =u=20m/s

So cosϕ=

20

10

=

2

1

⇒ϕ60

∘

Horizontal range =R=

g

u

2

sin2ϕ

=

10

20

2

sin120

∘

R=20

3

m

Maximum height =H=

2g

u

2

sin

2

ϕ

=

2×10g

20

2

×

4

3

H=15m