Question

the angle of projection for maximum horizontal range is a)90 b)45 c) 0 d)180​

Answer 1

Answer:

b. 45°

Explanation:

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Answer 2

Answer:

answer: b) 45

Explanation:

Let Velocity is V and the angle of projection with horizontal is θ.

∴Range, R=g2V 2

sinθcosθ

= gV 2

sin2θ

⇒ dθ

dR

= gV 2

2cos2θ=0 ⇒2θ=90

⇒θ=45

⇒ dθ 2

d 2R

=− gV 2

4sin2θ, at θ=45

o

⇒ dθ 2

d 2R

=− g4V 2<0

Hence, at θ=45 o , Range is maximum.

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