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Bas_tet [7]
3 years ago
14

A. With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 45m?

Physics
1 answer:
bazaltina [42]3 years ago
6 0

Answer:

A. 29.7 m/s

B. 6.06 s

Explanation:

From the question given above, the following data were obtained:

Maximum height (h) = 45 m

A. Determination of the initial velocity (u)

Maximum height (h) = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 m/s (at maximum height)

Initial velocity (u) =.?

v² = u² – 2gh (since the ball is going against gravity)

0² = u² – (2 × 9.8 × 45)

0 = u² – 882

Collect like terms

0 + 882 = u²

882 = u²

Take the square root of both side

u = √882

u = 29.7 m/s

Therefore, the ball must be thrown with a speed of 29.7 m/s.

B. Determination of the time spent by the ball in the air.

We'll begin by calculating the time take to reach the maximum height. This can be obtained as follow:

Maximum height (h) = 45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) to reach the maximum height =?

h = ½gt²

45 = ½ × 9. 8 × t²

45 = 4.9 × t²

Divide both side by 4.9

t² = 45/4.9

Take the square root of both side

t = √(45/4.9)

t = 3.03 s

Finally, we shall determine the time spent by the ball in the air. This can be obtained as follow:

Time (t) to reach the maximum height = 3.03 s

Time (T) spent by the ball in the air =?

T = 2t

T = 2 × 3.03

T = 6.06 s

Therefore, the ball spent 6.06 s in the air.

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Answer:

F = \frac{4kqQb}{(b^2 + \frac{d^2}{2})^{1.5}}

Explanation:

Since all the four charges are equidistant from the position of Q

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3 0
3 years ago
Find the mass. 10 points. Will give brainliest.
TiliK225 [7]

Answer:

<h2>3.94 kg</h2>

Explanation:

Given,

Force ( f ) = 30 N

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\frac{7.6 \: m}{7.6}  =  \frac{30}{7.6}

Calculate

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Hope this helps..

Best regards!!

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Hi pupil here's your answer ::

_____________________________

How does Newton's second law of motion gives the measurement of force?
So the answer is first : what is newton's second law? =》The rate of change of momentum of an object is equivalent to particular direction of the FORCE
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hope that it helps. . . . . .
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