Question

If R is the range of a projectile on a horizontal plane and , r/8 its maximum height, then maximum horizontal range with same speed of projection is

Answer 1

Answer:

R=

g

16sin2θ

h=

2g

16sin

2

θ

R

h

=

g

16sin

2

θ

×

16sinθ

g

=

4

1

tanθ

=tanθ=

R

4h

X=

R

2

+(4h)

2

sinθ=

R

2

+16h

2

4h

R

max

=

16h

2

2h(R

2

+16h

2

)

R

max

=

8h

R

2

+2h

solution

Answer 2

R= g16sin2θ

h= 2g16sin2θ

​Rh= g16sin 2 θ× 16sinθg = 4tan=tanθ= R4h

​

X= R 2+(4h) 2

sinθ= R 2 +16h 2

​

4h

​

R

max

​

=

16h

2

2h(R

2

+16h

2

)

​

R

max

​

=

8h

R

2

​

+2h