An outfielder throws the baseball to first base located 56m away from the fielder,with a velocity of 50m/s. At what launch angle above the horizontal should he throw the ball for the first baseman to catch the ball at the same height?

Question

agent0smith · Answer

This seems more of a physics problem than a maths problem...$$\Large R = \frac{ v^2 }{ g } \sin (2 \theta)$$
R is the range (56m), v = 50m/s, g is gravity, and theta is the angle.

anonymous · Answer

yes, it is a physics problem. so when solving for theta, what do i do with sin(2) then?

agent0smith · Answer

Solve the equation for 2(theta), then divide by 2 to get theta.

anonymous · Answer

ok,let me try that...what about the sin

agent0smith · Answer

Well I'm assuming you know how to solve it once you get sin(2theta) = ...

agent0smith · Answer

Use inverse sine

anonymous · Answer

yeah i should be able to solve it..trying right now

anonymous · Answer

i got 6.35, can you please counter check if I am right

agent0smith · Answer

That looks about right.

anonymous · Answer

thanks, i appreciate it...we been trying to solve this problem since yesterday...just decided to check the net for help....the answer have 6.2 and 6.3 as separate option..will go with 6.35

agent0smith · Answer

Yeah, I got 6.34 degrees, using g=9.8/s^2

anonymous · Answer

yeah that is right..thanks again