Well, how much projectile motion are you familiar with?
Can you split the velocity into x and y components?
Are you familiar with formulae such as v=u+at, v^2=u^2+2as or s = ut+1/2(at^2)
You have the quarterback at O(0,0) throwing the ball at 25m/s at an angle of 75°.
First, we need to split the velocity into x and y components.
Imagine you have a right angled triangle of hypotenuse of length 25 units, and θ = 75°
From this you can work out the other two side lengths, the vertical and the horizontal one as being 25sin75 and 25cos75 respectively.
Now, you can treat this as two seperate equations.
Generally, you assume that the only force acting upon the object is the force of gravity, with g=9.8m/s
So, we have a vertical speed of 25sin75, accelerating at -9.8m/s (gravity acts downwards).
Logically, the highest point is the one where is stops. Any longer, it's moving down again, any earlier, and it's still moving up.
We can use out equation of v=u+at. It's stopped, so u = 0. Now you can sub in values and rearrange for t, which is how long.
Now, you have a time, acceleration and initial velocity. Thus, to get the max height, we can use s=ut+1/2(at^2). Remembering that this gives us the height above the football field we can work out the height above the parking lot.
Now, you continue on to find when it hits the ground, sub in to get total time in the air, then distance travelled to see where you need to be to catch it.
Remember that we assume only force of gravity is acting, thus horizontal velocity is constant. Thus, with the time in the air, the range is given by d=vt.
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