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courtrigrad
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A football player throws a football with an initial upward velocity component of 18.0 m/s and a horizontal velocity component of 25.0 m/s. (a) How much time is required for the football to reach the highest point of the trajectory? (b) How high is this point? (c) How much time (after being thrown) is required for the football to return to its original level? How does this compare with the time calculated in part (a)? (d) How far as it traveled horizontally during this time?
(a) Ok so I know that [itex] v_{x}_{0} = 25.0 m/s, v_{y}_{0} = 18.0 m/s [/itex]. So for part (a) we know that the vertical component of velocity [itex] v_{y} = 0 [/itex]. So would I use [itex] t = \frac{v_{0}\sin\theta_{0} - v_{y}}{g} [/itex] where [itex] v_{y}_{0} = v_{0} = 18.0 m/s, v_{y} = 0, \theta = 90, g = 9.8 m/s^{2} [/itex]?
(b) Would I use [itex] y = (v_{0}\sin\theta_{0})t - \frac{1}{2}gt^{2} [/itex] and plug in the time?
(c) wouldn't the times for part(a) and part (c) be equal neglecting air resistance?
(d) I would just find the time for the football to travel the horizontal distance and use the equation [itex] R = (v_{0}\cos\theta_{0})t_{2} [/itex]?
Thanks
(a) Ok so I know that [itex] v_{x}_{0} = 25.0 m/s, v_{y}_{0} = 18.0 m/s [/itex]. So for part (a) we know that the vertical component of velocity [itex] v_{y} = 0 [/itex]. So would I use [itex] t = \frac{v_{0}\sin\theta_{0} - v_{y}}{g} [/itex] where [itex] v_{y}_{0} = v_{0} = 18.0 m/s, v_{y} = 0, \theta = 90, g = 9.8 m/s^{2} [/itex]?
(b) Would I use [itex] y = (v_{0}\sin\theta_{0})t - \frac{1}{2}gt^{2} [/itex] and plug in the time?
(c) wouldn't the times for part(a) and part (c) be equal neglecting air resistance?
(d) I would just find the time for the football to travel the horizontal distance and use the equation [itex] R = (v_{0}\cos\theta_{0})t_{2} [/itex]?
Thanks
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