The quarterback of a football team releases a pass at a heights of 7 feet above the playing field, and the football is caught at a height of 4 feet, 30 yards directly downfield. The pass is released at an angle of 35 degrees with the horizontal. The parametric equations for the path of the football are given by x= 0.82vot and y= 7+ 0.57vot - 16t^2 where vo is the speed of the football (in feet per second) when it is released. Find the speed when the football is released.
To find the speed (vo) at which the football is released, we need to use the given information about the parametric equations for the path of the football.
First, let's consider the equation for the vertical component of the path (y-direction).
We know that the football is released at a height of 7 feet and caught at a height of 4 feet. Taking the difference, we get:
4 feet - 7 feet = -3 feet
This tells us that the football falls 3 feet during its trajectory.
According to the equation for the y-direction (y= 7+ 0.57vot - 16t^2), the coefficient of t^2 is -16, which represents the acceleration due to gravity (g), as it acts in the direction opposite to the positive y-axis.
Since the football falls 3 feet, we can set y equal to -3 and solve for t:
-3 = 7 + 0.57vot - 16t^2
After rearranging this equation, we get:
16t^2 - 0.57vot + 10 = 0
Now, let's consider the equation for the horizontal component of the path (x-direction).
The equation for x (x= 0.82vot) tells us that the horizontal displacement of the football is given by 0.82vot.
We are given that the football is caught 30 yards downfield, which can be converted to feet by multiplying by 3 (since there are 3 feet in one yard):
30 yards * 3 = 90 feet
Setting x equal to 90, we can solve for t:
90 = 0.82vot
Now we have a system of equations:
16t^2 - 0.57vot + 10 = 0 ...(1)
0.82vot - 90 = 0 ...(2)
To solve this system of equations simultaneously, we can use substitution.
From equation (2), we can solve for t:
t = 90 / (0.82vo)
Now we can substitute this expression for t in equation (1):
16(90 / (0.82vo))^2 - 0.57vo(90 / (0.82vo)) + 10 = 0
Simplifying further:
16(90^2) / (0.82^2vo^2) - 0.57(90) / 0.82 + 10 = 0
Solving this equation will give us the value of vo, the speed at which the football is released. However, the calculation involved may require significant computational power.