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A quarterback claims that he can throw the football a horizontal distance of 183 m (200 yd). Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 30.0 above the horizontal. To evaluate this claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is launched and caught at the same vertical level and that air resistance can be ignored. For comparison, a baseball pitcher who can accurately throw a fastball at 45 m/s (100 mph) would be considered exceptional.
$v_{0}=45.5 \mathrm{m} \cdot \mathrm{s}^{-1}$
Physics 101 Mechanics
Chapter 3
Kinematics in Two Dimensions
Motion in 2d or 3d
Cornell University
University of Washington
Simon Fraser University
University of Winnipeg
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So the question states that a football player can throw a ball 183 meters, an angle of 30 degrees, and we're trying to find how fast he needs to throw the ball to make this happen. So the first thing we need to do is figure out how long the ball will be in the air. So to do this, we can separate our velocity into two components are horizontal component V of X and vertical component. View why? And we can find our V a. Y. Because we know that sign of 30 degrees is equal tow opposite overhype on news which is the of y over the so V of why will be equal to the times sign of 30 degrees. Now that we know this, we can use our Kinnah Matics equation which states that the change in displacement is equal to the initial velocity times Time was one have a T squared. So we know that the vertical displacement of the ball will be zero because it starts on the ground and it ends on the ground. So this will be zero. We know that the initial velocity in the vertical direction will be v sign 30 degrees. We're trying to find the time t We know that the acceleration will be equal the negative 9.8 meters per second squared because of gravity. And we have this t squared at the end. So we console for tea. Um, so we'll get V sign 30 degrees times t by striking that on both sides. Uh, it's equal to 1/2 negative. 9.8 t square and then when we saw for tea, we get that t is equal to 0.102 times are General Velocity V. And once again, I did this by dividing by 1/2 times negative 9.8 on both sides. And then I took this square root to get me team. So now that we know are t value, we can figure out what our velocity has to be because we know the distance. The ball travels 183 meters. We can take our initial horizontal Volesky V of X and plug it into this formula to give us, um what are Vima V Value must be so to find this v not in terms of, uh are to find this v not. We need to figure out what V of X is equal to. So we know the of X can be found because co sign of 30 degrees is equal to adjacent overhype on news of the of X over the so that means V of X is equal to be co sign of 30 degrees. So when we plug this in, we have Vico sign 30 degrees times our time, which is 0.102 times V, is equal to 183 meters. So now we can sell for what V is equal to. We can divide by 0.10 to co sign 30 degrees on both sides and then take the square root and that will give us V is equal to 45.5 meters per second.
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