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

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) (Show Source): You can put this solution on YOUR website!
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

:
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 =
t^2 = 4.875
t =
t = 2.165 second
then
2.165 * 32 ft/sec = 69.282 ft, (agrees with the graph)

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