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Rufina [12.5K]
3 years ago
12

A football is kicked straight up in the air , it hits the ground 5.6 s later. What was the greatest height kicked reached by the

ball? Assume it was kicked at ground level.
Physics
2 answers:
marin [14]3 years ago
5 0

Answer: 38.5 m

Explanation:

1) Flight time = 5.6 s (given)

2) Ascending time = fligth time / 2 = 5.6s / 2 = 2.8 s

3) Vf = Vo - gt

Hmax ⇒ Vf = 0

⇒ 0 = Vo - gt ⇒ Vo = gt = 9.81m/s² (2.8s) = 27.5 m/s

4) H = Vot - gt² / 2

⇒ H = 27.5m/s(2.8s) - 9.81m/s² (2.8s)² / 2 = 77m - 38.5m = 38.5m

Answer: 38.5m

Reil [10]3 years ago
3 0
We'll just consider the down trip, from when it was at its highest point to when it hits the ground.  Initial velocity is 0 m/s, acceleration is 9.81 m/s^2 (gravity), and time is 5.6s.

One of the kinematic equations is d = (Vi)(t) + 0.5(a)(t^2).
Plugging it in gives us d = 0 + (0.5)(9.81)(5.6^2) = 153.8208 m.

Hope this helped!
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Learning Goal:
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Answer:

A. U_0 = \dfrac{\epsilon_0 A V^2}{2d}

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U = \dfrac{QV}{2}

Its charge, Q, is related to its capacitance by Q=CV (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for U,

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A. Substituting for C in U,

U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. When the distance is 3d,

U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}

U_1 = \dfrac{\epsilon_0 A V^2}{6d}

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U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

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