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Question 201491This question is from textbook algebra 1 expressions,equations, and applications
: an astronaut on the earth practices jumping to the ground from the spaceship resting 3 meters above the ground. the initial upward velocity of the jump is 4 m/sec.
a. when will the astronaut be back at the same level s the jump?
b. at what time is the highest point reached?how high above the ground is that?
c.when does the astronaut reach the ground?
This question is from textbook algebra 1 expressions,equations, and applications
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! In metric units, g = 9.8 meters/sec/sec
Height as a function of time:
h(t) = -9.8t^2 + 4t + 3
Set h = 3 to find the times at 3 meters.
3 = -9.8t^2 + 4t + 3
9.8t^2 - 4t = 0
t = 0
t = 4/9.8 = ~ 0.408 seconds
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b) The easiest way to find this is divide the time back at 3 meters by 2, since he travels up to apogee in the same time it takes to get back to 3 meters.
t at max height = 0.204 seconds
Max height = h(0.204)
hmax = ~ 3.40816 meters above ground, not above where he jumped from
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Set h = 0 for the time at ground level (3 meters below the start point)
9.8t^2 - 4t - 3 = 0
t = 0.7938 seconds
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