A baseball leaves a pitcher\'s hand horizontally at a speed of 163 km/h. The dis

Question

A baseball leaves a pitcher's hand horizontally at a speed of 163 km/h. The distance to the batter is 18.3 m. Neglect air resistance.

(a) How long does it take for the ball to travel the first half of that distance?


(b) How long does it take for the ball to travel the second half of that distance?


(c) How far does the ball fall under gravity during the first half?



(d) How far does the ball fall under gravity during the second half?


(e) Why aren't the quantities in (c) and (d) equal?

1)The horizontal component of the velocity is decreasing.

2) The vertical component of the velocity is decreasing.

3) The horizontal component of the velocity is increasing.

4)The vertical component of the velocity is increasing.

Explanation / Answer

(a) first half of the distancce d = 18.3 m/ 2   = 9.15 m time t = d / v   = 9.15 m / (163*10^3 m / 60 *60 sec) t = 0.202 sec (b) second half of the distancce d = 18.3 m/ 2   = 9.15 m time t = d / v   = 9.15 m / (163*10^3 m / 60 *60 sec) t = 0.202 sec (c)under gravity initial velocity  (verticle )u = 0.0 m/s distance d = u*t + (1/2) at^2 Here t = 0.211 sec a = 9.8 m/s^2 d = (1/.2)*9.8*(0.211)^2    = 0.199 m (d)under gravity initial velocity  (verticle )u = 0.0 m/s distance d = u*t + (1/2) at^2 Here t = 0.202+0.202 sec   = 4.22 sec a = 9.8 m/s^2 d = (1/.2)*9.8*(0.404)^2    = 0.799 m I hope this wiil help u (b) second half of the distancce d = 18.3 m/ 2   = 9.15 m time t = d / v   = 9.15 m / (163*10^3 m / 60 *60 sec) t = 0.202 sec (c)under gravity initial velocity  (verticle )u = 0.0 m/s distance d = u*t + (1/2) at^2 Here t = 0.211 sec a = 9.8 m/s^2 d = (1/.2)*9.8*(0.211)^2    = 0.199 m (d)under gravity initial velocity  (verticle )u = 0.0 m/s distance d = u*t + (1/2) at^2 Here t = 0.202+0.202 sec   = 4.22 sec a = 9.8 m/s^2 d = (1/.2)*9.8*(0.404)^2    = 0.799 m I hope this wiil help u initial velocity  (verticle )u = 0.0 m/s distance d = u*t + (1/2) at^2 Here t = 0.202+0.202 sec   = 4.22 sec a = 9.8 m/s^2 d = (1/.2)*9.8*(0.404)^2    = 0.799 m I hope this wiil help u

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