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A ball is thrown from the top of a 36 m tall tower with an initial speed of 5 m/s at an angle of 37 degrees above the horizontal. Find the horizontal range of the ball measured from the base of the tower when it lands on the ground. (Take g = 10 m/s², sin 37 deg = 0.6, cos 37 deg = 0.8)

1. 12 m
2. 15 m
3. 18 m
4. 24 m

Correct answer: 12 m

Solution

With vx = 4 m/s and vy = 3 m/s, the vertical equation gives 5t² - 3t - 36 = 0, yielding t = 3 s. Horizontal range = 4 * 3 = 12 m.

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