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sergey [27]
3 years ago
5

A foul ball is hit straight up into the air with a speed of about 25 m/s. ? how high does it go? (b) how long is it in the air?

Physics
1 answer:
nikdorinn [45]3 years ago
5 0

To solve this problem, we make use of the equations of motion:

vf^2 = v0^2 – 2 g h

vf = v0 – g t

h = v0 t – 0.5 g t^2

 

A. Finding for max height:

Max height occurs when vf = 0, therefore:

0^2 = 25^2 – 2 (9.8) h

h = 31.89 m

 

B. Finding for t:

t = t(up) + t(down)

 

vf = v0 – g t(up)

0 = 25 – 9.8 t(up)

t(up) = 2.55 s

 

h = v0 t(down) – 0.5 g t(down)^2

31.89 = 0.5 (9.8) t(down)^2

t(down)^2 = 6.51

t(down) = 2.55 s

 

so,

<span>t = 5.1 s</span>

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Answer:

1. 24375 N/C

2. 2925 V

Explanation:

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E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}

E = 24375 N/C

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V = E x d

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8. Fig. 4.1 shows a heavy ball B of weight W suspended from a fixed beam by two ropes P and Q.
mart [117]

Answer:

The resultant tension of the two ropes is approximately 42.4 N

The length of the line representing the resultant tension is approximately 8.48 cm

Please find included  with the answer the scale drawing created with Microsoft Word

Explanation:

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The tension in rope P, T_P = 30 N

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The angle the rope, 'Q', makes with the horizontal = 45°

The scale factor of the scale diagram, S.F. = 5.0 N/cm

By the resolution of forces at equilibrium, we have;

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The length of the resultant force, W = W/(S.F.) ≈ 42.4 N/(5.0 N/cm) = 8.48 cm

The drawing of the vectors using the scale factor of 5.0 N/cm is created using Microsoft Word is included

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