A baseball leaves a pitcher\'s hand horizontally at a speed of159 km/h. The dist

Question

A baseball leaves a pitcher's hand horizontally at a speed of159 km/h. The distance to the batter is18.3 m. Neglect air resistance. (a) How long does it take for the ball totravel the first half of that distance?
s

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

(c) How far does the ball fall under gravity during the firsthalf?
m

(d) How far does the ball fall under gravity during the secondhalf?
m

(e) Why aren't the quantities in (c) and (d) equal? Thevertical component of the velocity is increasing. Thehorizontal component of the velocity isincreasing.     Thehorizontal component of the velocity is decreasing.Thevertical component of the velocity is decreasing. (a) How long does it take for the ball totravel the first half of that distance?
s

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

(c) How far does the ball fall under gravity during the firsthalf?
m

(d) How far does the ball fall under gravity during the secondhalf?
m

(e) Why aren't the quantities in (c) and (d) equal? Thevertical component of the velocity is increasing. Thehorizontal component of the velocity isincreasing.     Thehorizontal component of the velocity is decreasing.Thevertical component of the velocity is decreasing. Thevertical component of the velocity is increasing. Thehorizontal component of the velocity isincreasing.     Thehorizontal component of the velocity is decreasing.Thevertical component of the velocity is decreasing. Thevertical component of the velocity is increasing. Thehorizontal component of the velocity isincreasing.     Thehorizontal component of the velocity is decreasing.Thevertical component of the velocity is decreasing. Thevertical component of the velocity is increasing. Thehorizontal component of the velocity isincreasing.     Thehorizontal component of the velocity is decreasing.Thevertical component of the velocity is decreasing.

Explanation / Answer

159 km/h=44.167 m/s (a) the time  for the ball to travel the first halfof that distance t1=d/2v=0.207 s (b)the time needed=t1=0.207 s (c) the distance traveled when  the ball fall undergravity during the first half y1=g*t12/2=0.21 m (initialvertical velocity =0) (d) the distance traveled: y2=g*(2*t1)2/2 -y1=3*y1=0.63 m (e)The vertical component of the velocityis increasing (e)The vertical component of the velocityis increasing

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