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Olin [163]
3 years ago
5

A ball thrown horizontally at 12.6 m/s from the roof of a building lands 20.0 m from the base of the building

Physics
1 answer:
S_A_V [24]3 years ago
7 0

Answer:

1.59 seconds

12.3 meters

but if you are wise you will read the entire answer.

Explanation:

This is a good question -- if not a bit unusual. You should try and understand the details. It will come in handy.

Time

<u>Given</u>

a = 0 This is the critical point. There is no horizontal acceleration.

d = 20 m

v = 12.6 m/s

<u>Formula</u>

d = vi * t + 1/2at^2

<u>Solution</u>

Since the acceleration is 0, the formula reduces to

d = vi * t

20 = 12.6 * t

t = 20 / 12.6

t = 1.59 seconds.

It takes 1.59 seconds to hit the ground

Height of the building

<u>Givens</u>

t = 1.59 sec

vi = 0     Another critical point. The beginning speed vertically is 0

a = 9.8 m/s^2   The acceleration is vertical.

<u>Formula</u>

d = vi*t + 1/2 a t^2

<u>Solution</u>

d = 1/2 a*t^2

d = 1/2 * 9.8 * 1.59^2

d = 12.3 meters.

The two vi's are not to be confused. The horizontal vi is a number other other 0 (in this case 12.6 m/s horizontally)

The other vi is a vertical speed. It is 0.

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During lightning strikes from a cloud to the ground, currents as high as 2.50×104 A can occur and last for about 40.0 μs . How m
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<h3>The charge transferred from the cloud to earth is 1 Coulomb.</h3>

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From the formula of current,

<h3>  I = \frac{Q}{t}</h3>

Where Q = charge transfer between cloud and earth.

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the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

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3 years ago
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