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Papessa [141]
3 years ago
11

A ball is thrown horizontally from the top of a 65 m building and lands 115 m from the base of the building. Ignore air resistan

ce, and use a coordinate system whose origin is at the top of the building, with positive y upwards and positive x in the direction of the throw.
(a) How long is the ball in the air in seconds?
(b) What must have been the initial horizontal component of the velocity in m/s?
(c) What is the vertical component of the velocity just before the ball hits the ground in m/s?
(d) What is the magnitude of the velocity of the ball just before it hits the ground in m/s?
Physics
1 answer:
Mrrafil [7]3 years ago
4 0

Answer:

The answer to your question is:

a) t = 3.64 s

b) vox = 31.59 m/s

c) vy = 35.71 m/s

d) v = 47.67 m/s

Explanation:

a) To calculate the time, we know that voy = 0 m/s so we this this formula

   h = voy + 1/2(gt²)

   65  = 0 + 1/2(9.81)(t²)

   65 = 4.905t²

   t² = 65/4.905 = 13.25

   t = 3.64 s

b) To calculate vox  we use this formula vox = d/t ; vox is constant

   vox = 115/3.64 = 31.59 m/s

c) To calculate voy we use the formula    vy = voy + gt

but voy = 0

               vy = gt = 9.81 x 3.64 = 35.71 m/s

d) To calculate v  we use the pythagorean theorem

            c2 = a2 + b2

            c2 = 31.59² + 35.71² = 997.92 + 1275.20

            c2 = 2273.12

              c = 47.67 m/s

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