25 m/s. After 9. Pablo Sandoval throws a baseball with a horizontal component of

Question

25 m/s. After 9. Pablo Sandoval throws a baseball with a horizontal component of velocity of 2 seconds, the ball is 40 m above the release point. Calculate the horizontal distance it has traveled by this time, its initial vertical component of velocity. and its initial angle of projection. Also, is the ball on the way up or the way down at this moment in time? 17. In a a punter kicks the ball a horizontal distance of 43 yards (39 m). On TV football game From this information, calculate the they track the hang time, which reads 3.9 seconds. angle and speed at which the ball was kicked. (Note for non-football watchers: the of projectile starts and lands at the same height It goes 43 yards horizontally in a time 3.9 seconds)

Explanation / Answer

along horizontal

vox = vo*costheta = 25 m/s

acceleration ax = 0

horizontal distance x = vox*t = 25*2 = 50 m <<<------------answer

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along vertical

y = voy*t + (1/2)*ay*t^2

40 = v0y*2 - (1/2)*9.8*2^2

voy = 7.72 m/s <<<------------answer

----------------

angle theta = tan^-1(voy/vox)

theta = 17.16 <<<------------answer

-----------------

vy = voy + ay*t

vy = 7.72 - (9.8*2) = -11.88 m/s

vy is negative. so the ball is on the way down <<<------------answer

==================

17)

along vertical

total displacement y = 0

acceleration ay = -9.8 m/s^2

hang time = T

y = voy*T + (1/2)*ay*T^2

0 = voy*3.9 - (1/2)*9.8*3.9^2

voy = 19.11 m/s

along horizontal

acceleration ax = 0

distance x = 39 m

x = vox*t + (1/2)*ax*t^2

39 = vox*3.9

vox = 10 m/s

angle = tan^-1(voy/vox) = 62.3 degrees <<<<<<---------answer

speed vo = sqrt(vox^2+voy^2) = sqrt(19.11^2+10^2) = 21.56 m/s <<<---------answer

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