SOLUTION: A projectile is launched at an approximately 60-degree angle to the horizontal with an initial speed of 256 ft/sec. The equation for its vertical distance from the ground is h = 22
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Question 323520: A projectile is launched at an approximately 60-degree angle to the horizontal with an initial speed of 256 ft/sec. The equation for its vertical distance from the ground is h = 222t – 16t2, where t is time in seconds. Find the projectile’s maximum height and the total time that it is in the air. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A projectile is launched at an approximately 60-degree angle to the horizontal
with an initial speed of 256 ft/sec.
The equation for its vertical distance from the ground is h = 222t – 16t^2,
where t is time in seconds.
Find the projectile’s maximum height and the total time that it is in the air.
:
Find the axis symmetry of the equation h = -16t^2 + 222t
t =
t =
t = 6.9375 second for max height
Find the max height
h = -16(6.9375^2) + 222(6.9375)
h = -16(48.1289) + 1540.125
h = -770.0624 + 1540.125
h = +770.0625 ft max height
:
Total time in the air is when it hits the ground, then h=0
-16t^2 + 222t = 0
Divide equation by -16t
t - 13.875 = 0
t = 13.875 sec in the air
