Question

poora solve krke bhejna​

Answer 1

Here u=30ms  

−1

, Angle of projection, θ=90−30=60  

∘

 

Maximum height,

H=  

2g

u  

2

sin  

2

θ

​  

=  

2×10

30  

2

sin  

2

60  

∘

 

​  

=  

20

900

​  

×  

4

3

​  

=  

4

135

​  

m

Time of flight, T=  

g

2usinθ

​  

=  

10

2×30×sin60  

∘

 

​  

=3  

3

​  

s

Horizontal range = R=  

g

u  

2

sin2θ

​  

=  

10

30×30×2sin60  

∘

cos60  

∘

 

​  

=45  

3

​  

m