Every once in a while a bird will get hit by a baseball during a baseball game. The longest home runs in baseball land about 500 feet (which is 152.4 meters) from home plate. If these balls leave the bat at a 45 degree angle to maximize their distance, what is the safe height in meters for birds around baseball stadiums?
Details and assumptions

You may neglect air resistance during the flight of the ball
You may neglect the small initial height of the ball.
You may assume that gravity is a constant 9.81 m/s2.

Question

Every once in a while a bird will get hit by a baseball during a baseball game. The longest home runs in baseball land about 500 feet (which is 152.4 meters) from home plate. If these balls leave the bat at a 45 degree angle to maximize their distance, what is the safe height in meters for birds around baseball stadiums?
Details and assumptions

    You may neglect air resistance during the flight of the ball
    You may neglect the small initial height of the ball.
    You may assume that gravity is a constant 9.81 m/s2.

anonymous · Answer

the initial velocity of projectile=v, initial height=yi, gravitational acceleration=g at any time horizontal distance is x and vertical distance is y. maximum horizontal distance i.e. range=r. the angle of projection is=a so, the equations are, r=[v(cosa)/g][vsina+sqrt{(vsina)^2+2g(yi)}] and y=yi+xtana-[gx^2]/[2(vcosa)^2] now yi=0, and a=45. so, r=v^2/g and y=x-g(x/v)^2=x-x^2/r so to get maximum y, dy/dx=0 at x=xm. so, dy/dx=1-2x/r so, 1-2xm/r=0 =>xm=r/2 so, maximum y=ym=xm-(xm)^2/r=r/2-r/4=r/4 now, r=152.4 meter. so safe height=ym=r/4. you calculate the value. the site for the formulas, http://en.wikipedia.org/wiki/Trajectory_of_a_projectile