Football kicked 52 yards horizontally to a goal 10 ft high

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AI Thread Summary
A football is kicked horizontally towards a goal that is 10 feet high from a distance of 52 yards, with an initial velocity of 55 mph. The problem involves using kinematic equations to determine the angles at which the ball can be kicked to clear the goal. The horizontal velocity is converted to feet per second, and the horizontal and vertical components of motion are analyzed separately. The time of flight is expressed in terms of the angle theta, and the vertical displacement equation incorporates gravity to find possible values for theta. The next step involves substituting the time equation into the vertical motion equation to solve for the angles.
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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.
 
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