Question

what is the angle of projection at which the hmax and range are equal

Answer 1

ANSWER : tan∅ = 4

SOLUTION

Hmax = R

FORMULAS

H for projectile motion from ground to ground = u²sin²∅/2g

R for projectile motion from ground to ground = u²sin2∅/g

GIVEN: Hmax = R

∴u²sin²∅/2g = u²sin2∅/g

Cancel g and u on both sides

⇒sin²∅ = 2sin2∅

∵sin2x = 2sinxcosx

∴sin²∅ = 2×2sin∅cos∅

Cancel sin∅ both sides

⇒sin∅=4cos∅

⇒tan∅ = 4

I HOPE IT HELPS...!!!

Answer 2

Answer:

θ=tan  −1  (4)

Explanation:

Correct option is

C

θ=tan  

−1

4 or θ=70

R=  

g

u  

2

sin2θ

​  

 

H=  

2g

u  

2

sin  

2

θ

​  

 

g

u  

2

sin2θ

​  

=  

2g

u  

2

sin  

2

θ

​  

 

2sinθcosθ=  

2

sin  

2

θ

​  

 

ucosθ=sinθ

tanθ=4

θ=tan  

−1

4