Determine the maximum height and range of a projectile fired at a height of 4 feet above the ground with an initial velocity of 600 feet per second and at an angle of 45° above the horizontal. (round your answers to three decimal places.)

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

Ithink the answer would be 1,350!