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aliya0001 [1]
3 years ago
11

A ball is thrown directly downward with an initial speed of 7.95 m/s, from a height of 29.0 m. After what time interval does it

strike the ground?
Physics
1 answer:
Y_Kistochka [10]3 years ago
4 0

Answer: after 1.75 seconds

Explanation:

The only force acting on the ball is the gravitational force, so the acceleration will be:

a = -9.8 m/s^2

the velocity can be obtained by integrating over time:

v = -9.8m/s^2*t + v0

where v0 is the initial velocity; v0 = -7.95 m/s.

v = -9.8m/s^2*t - 7.95 m/s.

For the position we integrate again:

p = -4.9m/s^2*t^2 - 7.95 m/s*t + p0

where p0 is the initial position: p0 = 29m

p =  -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Now we want to find the time such that the position is equal to zero:

0 = -4.9m/s^2*t^2 - 7.95 m/s*t + 29m

Then we solve the Bhaskara's equation:

t = \frac{7.95 +- \sqrt{7.95^2 +4*4.9*29} }{-2*4.9} = \frac{7.95 +- 25.1}{9.8}

Then the solutions are:

t = (7.95 + 25.1)/(-9.8) = -3.37s

t = (7.95 - 25.1)/(-9.8) = 1.75s

We need the positive time, then the correct answer is 1.75s

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Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).

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An elevator motor provides 45.0 kW of power while lifting an elevator 35.0 m. If the elevator contains seven passengers each wit
mylen [45]

Find how much work ∆<em>W</em> is done by the motor in lifting the elevator:

<em>P</em> = ∆<em>W</em> / ∆<em>t</em>

where

• <em>P</em> = 45.0 kW = power provided by the motor

• ∆<em>W</em> = work done

• ∆<em>t</em> = 20.0 s = duration of time

Solve for ∆<em>W</em> :

∆<em>W</em> = <em>P</em> ∆<em>t</em> = (45.0 kW) (20.0 s) = 900 kJ

In other words, it requires 900 kJ of energy to lift the elevator and its passengers. The combined mass of the system is <em>M</em> = (<em>m</em> + 490.0) kg, where <em>m</em> is the mass of the elevator alone. Then

∆<em>W</em> = <em>M</em> <em>g h</em>

where

• <em>g</em> = 9.80 m/s² = acceleration due to gravity

• <em>h</em> = 35.0 m = distance covered by the elevator

Solve for <em>M</em>, then for <em>m</em> :

<em>M</em> = ∆<em>W</em> / (<em>g h</em>) = (900 kJ) / ((9.80 m/s²) (35.0 m)) ≈ 2623.91 kg

<em>m</em> = <em>M</em> - 490.0 kg ≈ 2133.91 kg ≈ 2130 kg

4 0
3 years ago
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