Asked by Anonymous
Sienna tosses a ball from the window of her high-rise apartment building with an initial velocity of 6.00 m/s at 28.0° above the horizontal. The ball strikes the ground 4.30 s later.
What is the horizontal distance between the base of the building and where the ball strikes the ground? What is the height it was thrown? What's the time it takes for the ball to reach the point 19.0 m below the window?
What is the horizontal distance between the base of the building and where the ball strikes the ground? What is the height it was thrown? What's the time it takes for the ball to reach the point 19.0 m below the window?
Answers
Answered by
oobleck
(a) 6.00 cos28° * 4.30 = _____
(b) h + 6.00 sin28° * 4.30 - 4.9 * 4.30^2 = 0
(c) 6.00 sin28° t - 4.9 t^2 = -19
(b) h + 6.00 sin28° * 4.30 - 4.9 * 4.30^2 = 0
(c) 6.00 sin28° t - 4.9 t^2 = -19
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