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Shalnov [3]
3 years ago
6

1. I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground

. Calculate the velocity in m/s after 1.10 seconds of freefall and calculate the velocity at impact in mi/hr.
Physics
2 answers:
kakasveta [241]3 years ago
7 0


what is the height of Fort Collins High School?

If found, use the formula, V=x/t-----------------(x=distance; t=time)

Svetradugi [14.3K]3 years ago
5 0
You have three known variables:

Acceleration - 9.8m/s^2
Time - 1.85s
Initial Velocity - 0m/s

For the first part of your question:

v = u + at

v = (0)+(9.8)(1.1)

v = 10.78 m/s  v=11 m/s (2 significant figures)

For the second part of your question:

v=u+at

v=(0)+(9.8)(1.85)
v=18.13 m/s



This still needs to be converted to m/h:

18.13m/s = 18.13\times 3600 metres/h=65628  metres/hour

65628 metres/hour = 65628\div1600 mi/h = 40.7925 mi/h

= 41 mi/hr (2 significant figures)





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Alice climbs 100 meters directly up the face of a cliff to reach the summit. Her friend Peter hikes the long way around taking a
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4 0
3 years ago
If the initial velocity of a ball is sent straight upward at 10.5m/s from the ground what will its final velocity be when it hit
Shalnov [3]

Answer: -10.08 m/s

Explanation:

Here we only need to analyze the vertical problem.

When the ball is in the air, the only force acting on it will be the gravitational force, this means that the acceleration of the ball, is equal to the gravitational acceleration, then:

a(t) = -9.8m/s^2

Where the negative sign is because gravity pulls the ball down.

To get the velocity equation we need to integrate over time, we get:

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

Where v0 is the initial vertical velocity, here it is v0 = 10.5 m/s

Then the velocity equation is:

v(t) =  (-9.8m/s^2)*t + 10.5 m/s

To get the position equation, we need to integrate again over time, we get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t + p0

Where p0 is the initial position, we know that the ball is sent upward from the ground, so p0 = 0m

Then the position equation is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

Now we need to find the value of t such that the position is equal to zero (this means that the ball hits the ground again).

Then we need to solve:

p(t) = 0 =  (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

If we divide both sides by t, we get:

0 =   (1/2)*(-9.8m/s^2)*t + (10.5 m/s)

Now we can solve it:

(1/2)*(9.8m/s^2)*t = 10.5 m/s

t = (10.5 m/s)/((1/2)*(9.8m/s^2)) = 2.14 s

This means that after 2.14 seconds, the ball will hit the ground again.

The velocity of the ball when it hits the ground is equal to:

v(2.14s) = (-9.8m/s^2)*2.14s + 10.5 m/s = -10.08 m/s

3 0
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