Physics: Projectile Motion - Maximizing Soccer Ball Distance/Hang-Time

[YMMMA]

TIme of Flight= 2v₀sinθ/g and Horizontal Range=v₀²sin2θ/g
Looking at both the equations, it's easily concluded that by increasing v₀ both Time of Flight and Horizontal Range will increase.
Not the tricky part is with θ.
For the given value of v₀, Horizontal Range at θ and 90°-θ are equal although the time of flight is different. Why?
Because in the equation of Horizontal Range, we have sin2θ and you can check that value of sin2θ for θ=30° and θ=60° are same. Hence if v₀ is also constant, then Horizontal Range for θ=30° and θ=60° will be same. So it's now clear that increasing θ will not increase Horizontal Range forever.
So out of the options present, I'll put my money on Option. C

You’re definitely right. I was finally able to recognize this at the end. Thank you all for your efforts...

 

[nrqed]

TIme of Flight= 2v₀sinθ/g and Horizontal Range=v₀²sin2θ/g
Looking at both the equations, it's easily concluded that by increasing v₀ both Time of Flight and Horizontal Range will increase.
Not the tricky part is with θ.
For the given value of v₀, Horizontal Range at θ and 90°-θ are equal although the time of flight is different. Why?
Because in the equation of Horizontal Range, we have sin2θ and you can check that value of sin2θ for θ=30° and θ=60° are same. Hence if v₀ is also constant, then Horizontal Range for θ=30° and θ=60° will be same. So it's now clear that increasing θ will not increase Horizontal Range forever.
So out of the options present, I'll put my money on Option. C

We much prefer to help students figure out the answers rather than providing them with a full solution.

By the way, I would have preferred for the OP to realize that when we plot $\sin(2 \theta)$ from 0 to 90 degrees (the range of possible angles $\theta$), the value of $\sin(2 \theta)$ increases as $\theta$ goes from 0 to 45 degrees but then actually decreases as $\theta$ goes from 45 to 90 degrees. So as long as the angle is below 45 degrees, we gain in range by increasing it, but past 45 degrees, we start losing in range. So the optimum angle for the range is 45 degrees. This is was I was aiming at with my guiding questions.

 

[YMMMA]

We much prefer to help students figure out the answers rather than providing them with a full solution.

You certainly made me able to figure out the solution. Thanks a million.

 

[Harenderjeet Arora]

We much prefer to help students figure out the answers rather than providing them with a full solution.

I am sorry, I didn't know. Just joined the Forum now, will keep that in mind.

 

[nrqed]

I am sorry, I didn't know. Just joined the Forum now, will keep that in mind.

No problem at all! Your help will be appreciated by many students I am sure. I just wanted to give you an idea of how we work here since you just joined us. And welcome!