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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

STEP 1

1. The motion of the football can be modeled using the equations of projectile motion.
2. The initial height of the football is 6.25 feet.
3. The initial velocity of the football is 73 feet per second.
4. The only force acting on the football after it is thrown is gravity, which accelerates it downward at 32 feet per second squared.

STEP 2

1. Identify the formula for projectile motion.
2. Substitute the given values into the formula.
3. Write the equation for the projectile motion.
4. Sketch the graph of the equation.

STEP 3

The formula for the vertical position y y of a projectile at time t t is given by:
y(t)=−12gt2+v0t+h0 y(t) = -\frac{1}{2} g t^2 + v_0 t + h_0
where g g is the acceleration due to gravity (32 ft/s2^2), v0 v_0 is the initial velocity, and h0 h_0 is the initial height.

STEP 4

Substitute the given values into the formula:
- g=32 g = 32 ft/s2^2 - v0=73 v_0 = 73 ft/s - h0=6.25 h_0 = 6.25 ft
y(t)=−12×32×t2+73×t+6.25 y(t) = -\frac{1}{2} \times 32 \times t^2 + 73 \times t + 6.25

STEP 5

Simplify the equation:
y(t)=−16t2+73t+6.25 y(t) = -16t^2 + 73t + 6.25
This is the equation modeling the projectile motion of the football.

STEP 6

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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