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Sagot :
Let distance between player A and B be x
from,
v = u + at
but,
v = u - gt, for the ball to be acted upon by gravity.
and the negative is to show the direction of g which is down.
u = initial speed(avg speed)
v will be 0m/s since it'll rest after reaching player B.
then,
0= u - (9.8*1.2)
0 =u - 11.76 m/s
u = 11.76 m/s
Therefore, the avg speed is 11.76m/s
Assuming one dimension & no gravity, friction, etc:
• Speed = distance/time
* A = (x1)
• B= (x2)
• Time = 1.2 sec
(|x1 - x2|)/1.2
Assuming two dimensions & no gravity, friction, etc:
• Speed = distance/time
* A = (x1, y1)
• B= (x2, y2)
• Time = 1.2 sec
(|x1 - x2| + |y1 - y2|)/1.2
Assuming 3 dimensions & no gravity, friction, etc.:
* Speed = distance/time
* A = (x1, y1, z1)
* B = (x2, y2, z2)
* Time = 1.2 sec
(|x1 - x2| + |y1 - y2| + |z1 - z2|)/1.2
• Speed = distance/time
* A = (x1)
• B= (x2)
• Time = 1.2 sec
(|x1 - x2|)/1.2
Assuming two dimensions & no gravity, friction, etc:
• Speed = distance/time
* A = (x1, y1)
• B= (x2, y2)
• Time = 1.2 sec
(|x1 - x2| + |y1 - y2|)/1.2
Assuming 3 dimensions & no gravity, friction, etc.:
* Speed = distance/time
* A = (x1, y1, z1)
* B = (x2, y2, z2)
* Time = 1.2 sec
(|x1 - x2| + |y1 - y2| + |z1 - z2|)/1.2
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