Math  /  Calculus

QuestionDuring a local high school foothall game, the quarterback for the home team attempts a deep pass to his wide recelver. The ball is launched from 6.25 feet with an initial velocity of 73 feet per second (f/s). a) Write the equation modeling the projectile motion for the football and sketch the graph.

Studdy Solution
To sketch the graph of the equation y(t)=16t2+73t+6.25 y(t) = -16t^2 + 73t + 6.25 :
- Recognize that this is a quadratic equation, which graphs as a parabola. - The parabola opens downward because the coefficient of t2 t^2 is negative. - The initial height of the football is 6.25 feet, which is the y-intercept of the graph. - The vertex of the parabola can be found using the vertex formula t=b2a t = -\frac{b}{2a} where a=16 a = -16 and b=73 b = 73 .
Calculate the vertex:
t=732×16=7332 t = -\frac{73}{2 \times -16} = \frac{73}{32}
- Substitute t=7332 t = \frac{73}{32} back into the equation to find the maximum height.
The graph should depict a downward-opening parabola starting at y=6.25 y = 6.25 and peaking at the calculated vertex.
The equation for the projectile motion is:
y(t)=16t2+73t+6.25 y(t) = -16t^2 + 73t + 6.25

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