SOLUTION: Consider the path of a projectile projected horizontally with a velocity of v feet per second at a height of s feet, where the model for the path is y=(-16/v^2)t^2 + s. (Air resist

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Consider the path of a projectile projected horizontally with a velocity of v feet per second at a height of s feet, where the model for the path is y=(-16/v^2)t^2 + s. (Air resist      Log On


   

Question 740135: Consider the path of a projectile projected horizontally with a velocity of v feet per second at a height of s feet, where the model for the path is y=(-16/v^2)t^2 + s. (Air resistance is disregarded and y is the height (in feet) of the projectile t seconds after its release). A ball is thrown from the top of a 75-feet tower with a velocity of 32 feet per second.
a). Find the equation of the parabolic path.
b). How far does the ball travel horizontally before striking the ground?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Consider the path of a projectile projected horizontally with a velocity of v feet per second at a height of s feet, where the model for the path is y=(-16/v^2)t^2 + s.
(Air resistance is disregarded and y is the height (in feet) of the projectile t seconds after its release).
A ball is thrown from the top of a 75-feet tower with a velocity of 32 feet per second.
a). Find the equation of the parabolic path.
y = (-16/32^2)t^2 + 75
looks like this
+graph%28+300%2C+200%2C+-10%2C+80%2C+-20%2C+120%2C+%28-16%2F32%5E2%29x%5E2%2B75%29+
:
b). How far does the ball travel horizontally before striking the ground?
Determine how long the ball is in the air
Determined by gravity -16t^2
-16t^2 + 75 = 0
-16t^2 = -75
t^2 = %28-75%29%2F%28-16%29
t^2 = 4.875
t = sqrt%284.6875%29
t = 2.165 second
then
2.165 * 32 ft/sec = 69.282 ft, (agrees with the graph)