IW03- 1D Motion Begin Date:5/22/2018 12:00:00 AM- -Due Date: 5/29/2018 1:30:00 P

Question

IW03- 1D Motion Begin Date:5/22/2018 12:00:00 AM- -Due Date: 5/29/2018 1:30:00 PM End Date: 6/24/2018 11:59:00 PM (14%) Problem, 10: A dime is dropped from, a hot-air balloon that is 295 m above the ground and rising vertically at 9.5 m/s. For this problem use a coordinate system in which up is positive. ?25% Part (a) Find the maximum height, in meters, that the dime attains. 25% Part (b) Find its height above the ground, in meters, 4.00 s after being released 25% Part (c) Find its velocity, in meters per second, 4.00 s after being released. 25% Part (d) Find the time, in seconds, that it takes the dime to hit the ground. Grade Summary Potential 100 | | ! (r) 1718| 91 Submissions Attempts remaining (4% per attempt) sino cos() tan() atanO acotan0sinh0 coshO tanh0 cotanh0o ODegrees Radians Submit Hint I give up! Hints: 2% Feedback: 2% deduction per feedback. All O 2018 TA, LLC

Explanation / Answer

Solution:

a)Using kinamatics,

v² = v?² + 2as
0² = 9.5² + 2(-9.81)s
s = 4.6 m

so max height is

295 + 4.6 = 299.599 = 300m above ground.

b)The general equation for position given a constant acceleration is

s = s? + v?t + ½at²

If ground level is considered the origin with up as the positive direction,

s? = 295 m
v? = 9.5 m/s
a = g = -9.81 m/s²

s = 295 + 9.5(4.0) + ½(-9.81)4.0²

s = 254.52m.

c) v = v? + at
v = 9.5 - 9.81(4.0)

v = -29.77 m/s.

d)0 = 295 + 9.5(t) + ½(-9.81)t²

quadratic formula

t = -6.85 s (which we ignore as it occurred before the coin is dropped)
or
t = 8.784 s

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