Question

24- For same maximum height, the two angles of projection are
0 & 909
O ° &
30 ° & 60°
45 & 45°
SIR
None of these​

Answer 1

Answer:

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Explanation:

Let the two angles be  θ  

1

​

  and  θ  

2

​

 and the projectile velocity be u

Horizontal range        R=  

g

u  

2

sin(2θ  

1

​

)

​

=  

g

u  

2

sin(2θ  

2

​

)

​

=  

g

u  

2

sin(180−2θ  

2

​

)

​

 

⟹    2θ  

1

​

=180  

o

−2θ  

2

​

 

⟹          θ  

2

​

=90−θ  

1

​

                       

Maximum height attained              H  

1

​

=  

2g

u  

2

sin  

2

θ  

1

​

 

​

         

Similarly         H  

2

​

=  

2g

u  

2

sin  

2

θ  

2

​

 

​

=  

2g

u  

2

sin  

2

(90−θ  

1

​

)

​

=  

2g

u  

2

cos  

2

θ  

1

​

 

​

         

 

∴    H=H  

1

​

+H  

2

​

=  

2g

u  

2

(sin  

2

θ  

1

​

+cos  

2

θ  

2

​

)

​

=  

2g

u  

2

 

​