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# Problem 1

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## darenzwong chiles

on 6 October 2013

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#### Transcript of Problem 1

Problem 1
If Michael Jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang time (total time to move upwards to the peak and then return to the ground)?
Problem 2
A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).
Problem 3
A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one-half the total hang-time.)
Problem 10
With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.
Given:
a = -9.8 m/s^2
vf = 0 m/s
d = 1.29 m

Find:
vi = ?
t = ?
Problem 1
Solution:
for vi,
(0 m/s)^2 = vi^2 + 2(-9.8m/s^2)(1.29m)
vi^2 = 25.28 m^2/s^2
vi = 5.03 m/s
for t,
vf = vi + at
0 m/s = 5.03 m/s + (-9.8m/s^2)t
t = (5.03m/s)/(-9.8m/s^2)
t = 0.513 s
Given:
vi = 0 m/s
vf = 521 m/s
d = 0.840 m

Find:
a = ?
Problem 2
Solution:
for a,
(521 m/s)^2 = (0 m/s)^2 + 2a(0.840 m)
a = (271441 m^2/s^2)/(1.68 m)
a = 1.62 x10^5 m/s^2

Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
The observation deck of tall skyscraper 370 m above the street. Determine the time required for a penny to free fall from the deck to the street below.
A bullet is moving at a speed of 367 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)

A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.
It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 m/s2, determine the speed of the Jaguar before it began to skid.
A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.
A dragster accelerates to a speed of 112 m/s over a distance of 398 m. Determine the acceleration (assume uniform) of the dragster.

Problem 4
Solution:
d = vit + 0.5(at^2)
-370 m = (0 m/s )t + 0.5(-9.8m/s^2)t^2
t^2 = (-370 m)/(-4.9 m/s^2
t^2 = 75.5 s^2
t = 8.69 s
Problem 3
Solution:
Use first, vf = vi + at
0 m/s = vi + (-9.8 m/s^2)(3.13 s)
vi = 30.6 m/s
Now use, vf^2 = vi^2 + 2ad
(0 m/s)^2 = (30.6 m/s)^2 + 2(-9.8m/s^2)d
d = (-938 m/s )/(-19.6 m/s^2)
d = 47.9 m
Given:
vi = 0 m/s
d = -370 m
a = -9.8 m/s^2
Find:
t = ?
Problem 5
Solution:
(0 m/s^2) = (367 m/s)^2 + 2a(0.0621 m)
a = (-134689 m^2/s^2)/(0.1242 m)
a = -1.08 x 10^6 m/s^2
Given:
vi = 367 m/s
vf = 0 m/s
d = 0.0621 m
Find:
a = ?
Problem 6
Solution:
d = vit + 0.5a(t^2)
d = (0 m/s)(3.41 s) + 0.5(-9.8 m/s^2)(3.41 s)^2
d = 0 + 0.5(9.8 m/s^2)(11.63 s^2)
d =-57.0 m
Given:
a = -9.8 m/s^2
t = 3.41 s
vi = 0 m/s

Find:
d = ?
Given:
a = -3.90 m/s^2
vf = 0 m/s
d = 290 m
Find:
vi = ?
Problem 7
Solution:
(0 m/s)^2 = vi^2 + 2(-3.90m/s^2)(290 m)
2262 m^2/s^2 = vi^2
vi = 47.6 m/s
Given:
vi = 0 m/s
vf = 88.3 m/s
d = 1365 m
Find:
a = ?
t = ?
Problem 8
Solution:
for a, vf^22 = vi^2 +2ad
(88.3 m/s)^2 = (0 m/s)^2 + 2a(1365 m)
(2730 m)a = 7797 m^2/s^2
a = (7797 m^2/s^2)/(2730 m)
a = 2.86 m/s^2
for, t vf = vi + at
88.3 m/s^2 = 0 m/s + (2.86 m/s^2)t
t = (88.3 m/s)/(2.86m/s^2)
t = 30.8 s
Given:
vi = 0 m/s
vf = 112 m/s
d = 398 m
Find:
a = ?
Problem 9
Solution:
(112 m/s)^2 = (0 m/s)^2 + 2a(398 m)
12544 m^2/s^2= a(796 m)
a = (12544 m^2/s^2)/(796 m)
a = 15.8 m/s^2
Given:
a = -9.8 m/s^2
vf = 0 m/s
d = 91.5 m

Find:
vi = ?
t = ?

Problem 10
Solution:
First, Find speed, vf^2 = vi^2 +2ad
(0m/s)^2 = vi^2 + 2(-9.8 m/s^2)(91.5 m)
vi^2 = 1793 m^2/s^2
vi = 42.3 m/s

Now, convert from m/s to mi/hr
vi = 42.3 m/s(2.23 mi/hr)/(1 m/s)
vi =94.4 mi/hr
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