Use the model for projectile motion, assuming there is no air resistance. Determ
ID: 2857043 • Letter: U
Question
Use the model for projectile motion, assuming there is no air resistance.
Determine the maximum height and range of a projectile fired at a height of 4 feet above the ground with an initial velocity of 900 feet per second and at an angle of 45° above the horizontal. (Round your answers to three decimal places.)
Explanation / Answer
The sine and cosine of a 45 degree angle is 0.707.
The initial vertical velocity and the horizontal velocity will be 900 * 0.707 =636.3feet /sec
The maximum distance for the projectile to reach the firing elevation (4 feet above ground) will be
X = (v^2 / 16) * sine(2 * 45 degrees)
X = 900^2 /16* Sine(90 degrees)
X = 50625 feet
However, this is not quite the maximum range. See the final maximum range below.
The maximum height will be
h =( (v^2 / 2 *16) * Sine^2(45 degrees)) +4
h=((900^2 / 32) * 0.707^2) + 4
h =(3826.97) + 4
h = 12656.427 feet
The time to travel the initial 50625 feet will be 50625 / 636.3 = 80.3seconds.
It will take 80.3 / 2 = 40.15 seconds to reach maximum height.
The time to fall the 12656.427 feet to the ground will be
s = 4 t^2
12656.427 = 4t^2
t^2 = 12656.427 / 4
t=Sqrt(3164.106)
t = 56.25038 seconds
The total flight time of the projectile will be 40.15 + 56.25038= 96.4003 seconds.
The final maximum distance will be 96.4003 * 636.3= 60790.083feet.
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