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A projectile is fired vertically upward from a height of 200 feet above the ground, with an initial velocity of 900 ft/sec. Recall that projectiles are modeled by the function h(t)=−16t2+v0t+y0. Write a quadratic equation to model the projectile's height h(t) in feet above the ground after t seconds.

User Melsauce
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Answer:

The equation that model the projectile's height is
h(t) = -16.087\cdot t^(2)+900\cdot t +200

Explanation:

We know that projectile is fired vertically upward from a height of 200 feet above the ground with an initial velocity of 900 feet per second and projectiles experiment a parabolic motion, which combines horizontal uniform motion and vertical uniform accelerated motion due to gravity, whose equation of motion is:


h(t) = (1)/(2)\cdot g \cdot t^(2)+v_(o)\cdot t + y_(o)

Where:


h(t) - Current height above the ground at instant t, measured in feet.


y_(o) - Initial height, measured in feet.


v_(o) - Initial velocity, measured in feet per second.


g - Gravitational acceleration, measured in feet per square second.


t - Time, measured in seconds.

If we know that
g = -32.174\,(ft)/(s^(2)),
v_(o) = 900\,(ft)/(s) and
y_(o) = 200\,ft, the equation that model the projectile's height is


h(t) = -16.087\cdot t^(2)+900\cdot t +200

User Ramy Feteha
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