Question

show that maximum horizamtal range of an oblique projectile one fourt of its maximum height​

Answer 1

Answer:

We know that,

The range of the projectile R=

g

u

2

sin2θ

,R will be the maximum, if sin2θ=1⇒2θ=90

0

⇒θ=45

0

Then,

R

max

=

g

u

2

So the maximum height attained by the projectile is:

H=

2g

u

2

sin

2

45

0

=

g

u

2

×

2

1

=

4

R

max

⇒R

max

=4H

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Answer 2

Answer:MAXIMUM HEIGHT IS 1/4TH OF THE RANGE ONLY WHEN THE ANGLE OF PROJECTION IS 45°.

Explanation: