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devlian [24]
3 years ago
5

A golf ball is released from rest from the top of a very tall building. Choose a coordinate system whose origin is at the starti

ng point of the ball, and whose y axis points vertically upward. Neglecting air resistance, calculate the ve- locity of the ball after 3.04 s. The acceleration of gravity is 9.8 m/s2 . Answer in units of m/s.
Physics
2 answers:
Alinara [238K]3 years ago
7 0

Answer: v=29.792m.s^{-1}

Explanation:

<u>Here given that:</u>

  • initial velocity of the golf ball, u =0m.s^{-1}
  • a coordinate system pointing upwards with origin at the point of release.
  • time, t = 3.04 seconds
  • acceleration due to gravity, g = 9.8 m.s^{-2}

Now, we need to find the velocity (v) after the time 3.04 s.

From the given and required data we choose the suitable equation of motion: v = u+at

Putting the given values in the above equation:

v= 0 - gt ∵the given frame of reference points in the upward directoin so we have negative on the downward side.

v=0-9.8\times 3.04

v=29.792m.s^{-1}

Zigmanuir [339]3 years ago
5 0

Answer:

Velocity of the ball after 3.04 (s) = 29.79 (m/s)

Explanation:

From the free fall movement we have the following formulas: Vf^{2} = Vo^{2} - 2gh and h=Vo*t - \frac{g*t^{2} }{2}, First we need to find the height to time iqual to 3.04 s using the formula: h=Vo*t - \frac{g*t^{2} }{2} and remember that golf ball was released from the rest (Vo= 0 (m/s)) so h= (0 (m/s))*(3.04 (s)) - \frac{9.8 (m/s^2)*(3.04 (s))^{2} }{2}, we get: h = -45.28 (m) with the height that we have got, now the velocity of the ball is calculate using Vf^{2} = Vo^{2} - 2gh solving for Vf, we get: Vf = \sqrt{Vo^{2}-2*g*h } replacing the values given Vf = \sqrt{(0 m/s)^{2}-2*(9.8 m/s^2)*(-45.28 m) }, so we get: Vf = 29.79 (m/s).

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