Use g = 10 m/s2 in this problem to simplify the calculations. You climb 11.0 met

Question

Use g = 10 m/s2 in this problem to simplify the calculations.
You climb 11.0 meters up the mast of a tall ship. The ship itself is traveling with a constant velocity of 4.00 m/s east, and the water is calm so the mast remains vertical. When you have climbed the 11.0 m up the mast and are at rest with respect to the mast, you release two balls. The first ball you simply release from rest, relative to you. The second ball you give an initial velocity of 1.10 m/s, directed east, relative to you. Assume that air resistance is negligible.



How far from the base of the mast does the first ball land? Use a positive sign if you think the ball lands some distance east of the mast, and a negative sign if you think the ball lands some distance west of the mast.
__________meters

Explanation / Answer

The relative motion of the mast does not affect the projectile range so,
initial horizontal velocity is 1.10 m/s
Time of flight is given by

s = vy0t + 0.5 * - a * t2

- 11m = (0)t + 0.5 (- 10)t2 -- --------> since vertical initial velocity is zero.

- 11m = - 5 t2

   t = ± 11/ 5

We will take only the positive value.

Horizontal distance is given by

S = vt

   = 1.1m/s * (11/ 5)

   = 1.1 * 1.48 s

s = + 1.628 meter to the east.

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