Football kicked 52 yards horizontally to a goal 10 ft high

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SUMMARY

The problem involves calculating the angles at which a football, kicked at an initial velocity of 55 mph (equivalent to 242/3 ft/s), can clear a goalpost 10 ft high from a horizontal distance of 52 yards (156 ft). The kinematic equations are applied to separate the motion into horizontal and vertical components. The time of flight is expressed as t = 1.938/cos(θ), and the vertical displacement equation is used to find the relationship between the angle θ and the height of the ball. By substituting the time equation into the vertical displacement equation, one can derive the possible angles for θ that allow the ball to clear the goal.

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  • Kinematic equations of motion
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  • Conversion of units (mph to ft/s)
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Homework Statement


A football is kicked towards a goal (The little fork thingies, whetever they're called) 10 ft high. The horizontal distance is 52 yards. The initial velocity is 55 mph. Find all angles you can kick the ball at so that it goes over the goal.


Homework Equations


The kinematics equations
d=vit+(at^2)/2
d=(vi+vf)*t/2
Vf=Vi+at
Vf^2=Vi^2+2ad
f=ma
d=vt

The Attempt at a Solution


55 mph is 242/3 ft/s.
so i drew the triangle, with a horizontal length of 156 ft. The height is 10ft. I know i need to find the height in terms of theta.
I split this problem into the x and y directions.
for x, i know vf and vi are both cos(theta)242/3
d=156.
I used d=(vi+vf)*t/2.
it's basically d=cos(theta)242/3 * t
so i foudn that t=1.938/cos(theta)
and then I'm stuck. what do i do after that?
 
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Ok so you have,

<br /> t=\frac{1.938}{cos(\theta)}<br />

in the x direction.

You also know in the y-direction,

<br /> \Delta y = \frac{243}{3}sin(\theta) t - \frac{1}{2} g t^2<br />

You know what \Delta y is so substitute the two equations and solve for possible values of theta.
 
Last edited:

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