maximum height = 2801.290 ft
Maximum range = 11193.158 ft
First, calculate the horizontal and vertical velocities of the projectile.
v = sin(45°)*600 ft/s = 0.707106781 * 600 ft/s = 424.2640687 ft/s
h = cos(45°)*600 ft/s = 0.707106781 * 600 ft/s = 424.2640687 ft/s
Now the altitude of the shell is described by the equation
d = 4 ft + vT - 0.5AT^2
where
d = distance
v = initial velocity
A = acceleration due to gravity (32.174 ft/s^2)
T = time
The projectile will reach its maximum height just as it's vertical velocity reaches 0. So
T = 424.2640687 ft/s / 32.174 ft/s^2 = 13.18655028 s
Plugging in the known values into the formula gives
d = 4 ft+ (424.2640687 ft/s)13.18655028 s - 0.5*32.174 ft/s^2 *
(13.18655028 s)^2
d = 5598.579474 ft - 16.087 ft/s^2 * 173.8851083 s^2
d = 5598.579474 ft - 2797.289737 ft
d = 2801.289737 ft
d = 2801.290 ft
So the projectile will reach a maximum height of 2801.290 ft
The maximum range will happen when 4 + vT - 0.5AT^2 has a value of 0. So substitute the known values. Getting
0 = 4 + 424.2640687 T - 0.5* 32.174 T^2
0 = 4 + 424.2640687 T - 16.087 T^2
We now have a quadratic equation, use the quadratic formula to solve. a = -16.087, b = 424.2640687, c = 4
T is either -0.009424722 seconds, or 26.38252528 seconds.
So the projectile will travel for 26.38252528 seconds before hitting the ground. Just multiply that time by the velocity.
26.38252528 s * 424.2640687 ft/s = 11193.15752 ft
Rounding to 3 decimal places gives 11193.158 ft
