Sorry this is such a long answer, but it's the nature of the beast.
The angle that yields the maximum possible range is #45^@#. Lets say a projectile was launched with an initial velocity of #25.00"m/s"# at #45^@# above the horizontal. What is the horizontal displacement and the maximum height reached by the projectile? What is their relationship?
Known/Given/Calculated:
initial velocity:#v_i=25.00"m/s"#
horizontal acceleration: #a_x=0"m/s"#
vertical acceleration: #a_y=g=-9.81"m/s"^2"#
angle, #theta##=##45^@#
initial horizontal velocity:
#v_"ix"=v_icostheta=25.00"m/s"*cos45^"o"=17.68"m/s"#
initial vertical velocity:
#v_"iy"=v_isintheta=25.00"m/s"*sin45^"o"=17.68"m/s"#
final vertical velocity:
#v_"fy"=-17.68"m/s"# (due to the symmetrical nature of the trajectory)
final horizontal velocity:
#v_"fx"=17.68"m/s"# (due to horizontal acceleration of 0m/s)
Equations:
final vertical velocity: #v_"fy"=v_"iy" + a_y*t#, where #a_y# is acceleration due to gravity, #-9.81"m/s"^2#.
horizontal displacement: #x=v_"ix"t+0.5a_xt^2#
vertical displacement: #y=v_"iy"*t+0.5a_y*t^2#
Solution:
Find the time using the equation for final vertical velocity.
#t=(v_"fy"-v_"iy")/(g)=(-17.68"m/s"-17.68"m/s")/(-9.81"m/s"^2)#
#t=(-35.36"m/s")/(-9.81"m/s"^2)=3.60"s"#
Find the horizontal displacement, #x#.
#x=v_"ix"t+0.50a_xt^2=v_"ix"t+0#
#x=v_"ix"t=17.68"m/s"*3.60"s"=63.6"m/s"#
Find the maximum vertical displacement (height), #y_"max"#.
Since maximum vertical displacement occurs halfway through the trajectory, we have to divide the time in half. So for maximum height, the time will be #1.80"s"#.
#y_"max"=v_"iy"t+0.5*a_y*t^2#
#17.68"m/s"xx1.80"s"+0.5xx-9.81"m/s"^2xx(1.80"s")^2=15.9"m"#
Now, divide #x# by #y_"max"#.
#x/y_"max"=(63.6"m/s")/(15.9"m/s")=4# exactly. So the horizontal range is 4 times the maximum height attained by the projectile.
