Asked by #1
A ball with mass 0.8kg is thrown southward into the air with a speed of 30m/s at an angle of 30 to the ground. A west wind applies a steady force of 4N N to the ball in an easterly direction. Where does the ball land and with what speed?
Answers
Answered by
oobleck
how long does the ball stay in the air?
y = 30(1/2)t - 4.9t^2 = 0
t = 1.75 seconds
The horizontal component of the speed is
30(.866) m/s south, meaning the ball goes 45.5 m south
eastward, a = F/m = 4/.8 = 5 m/s^2
so the distance eastward is 1/2 at^2 = 2.5 * 1.75^2 = 7.65m east
the speed east is v = at = 5*1.75 = 8.75 m/s
so the final speed at impact is √(8.75^2 + 25.98^2) = 27.4 m/s
y = 30(1/2)t - 4.9t^2 = 0
t = 1.75 seconds
The horizontal component of the speed is
30(.866) m/s south, meaning the ball goes 45.5 m south
eastward, a = F/m = 4/.8 = 5 m/s^2
so the distance eastward is 1/2 at^2 = 2.5 * 1.75^2 = 7.65m east
the speed east is v = at = 5*1.75 = 8.75 m/s
so the final speed at impact is √(8.75^2 + 25.98^2) = 27.4 m/s
Answered by
Anonymous
Everybody that have tried solving this problem gives me a different answer. I wish there is answer key so I know who are correct and who are not.
THANK oobleck
THANK oobleck
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