A quarterback claims that he can throw the football a horizontal distance of 196

Question

A quarterback claims that he can throw the football a horizontal distance of 196.6 m (215 yd). Furthermore, he claims that he can do this by launching the ball at the relatively low angle of 35° above the horizontal. To evaluate his 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. Answer in m/s.

Explanation / Answer

1) Determine Hang Time Hang time depends on the launch speed, the launch angle, and g, the acceleration due to gravity. Hang time only depends on motion in the y-direction: ?y = 0, assuming cannonball falls back down to same level a = -g vi = vsin? ?y = vi(t) + ½a(t)² 0 = v(sin?)(t) + ½(-g)(t)² 0 = t(vsin? - ½gt) Clearly, there are two points in time when the projectile is at ground level. t = 0, is the time when the projectile is launched. let 0 = vsin? - ½gt to find out when it lands ½gt = vsin? t = 2vsin?/g This result can also be obtained by showing that the initial vertical speed is equal to the final vertical speed but opposite in direction: vf² = vi² + 2g?y since change in height is 0, vf² = vi² vf = -vi, since the projectile moves up on the way up and down on the way down Using: a = (vf - vi)/t t = (vf - vi)/a t = (-vi - vi)/a t = -2vi/a t = -2vsin?/-g t = 2vsin?/g 2) Determine Range To determine range, analyse motion in the horizontal or x-direction: ?x = vt, where v is the constant horizontal speed of the projectile ?x = (vcos?)(2vsin?/g) ?x = v²2cos?sin?/g Using the identity: sin(2?) = 2cos?sin?, ?x = v²sin(2?)/g v is launch speed ? is launch angle g is acceleration due to gravity (9.8 m/s²) 3) Determine Launch Speed v = sqrt[?xg/sin(2?)] v = 45 m/s

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