Question

At what angle of projection the horizontal range is equal to maximum height?​

Answer 1

Answer:

the maximum range for projectile motion (with no air resistance) is 45 degrees.

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Explanation:

R= u²sin2θ/g

H=u² sin²θ/2g

u² sin2θ/g= u²sin²θ/2g

2sinθcosθ= sin² θ/2

4cosθ=sinθ

tanθ=4

Answer 2

Answer:

A) The maximum range is 4 times the maximum height attained during its flight in the case of a projectile. B) In the case of a projectile the range R is related to the time of flight T as R=5T2. If g=10 ms−2 the angle of projection is 45o.

Explanation:

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