SOLUTION: This is a word problem.
#116 Time of flight. Use the information from exercise 115 to determine how long Johnson was in the air. For how long was he more than 14ft in the air.
Ex
Question 81936This question is from textbook
: This is a word problem.
#116 Time of flight. Use the information from exercise 115 to determine how long Johnson was in the air. For how long was he more than 14ft in the air.
Exersice 115
For 1989 and 1990 Dave Johnson had the highest decathlon score in the world. When Johnson reached a speed of 32 ft/sec on the pole vault runway, his height above the ground t seconds after leaving the ground was given by h=-16t^2+32t.
How do I solve Problem 116? Please help This question is from textbook
You can put this solution on YOUR website! #116 Time of flight. Use the information from exercise 115 to determine how long Johnson was in the air. For how long was he more than 14ft in the air.
Exercise 115
For 1989 and 1990 Dave Johnson had the highest decathlon score in the world. When Johnson reached a speed of 32 ft/sec on the pole vault runway, his height above the ground t seconds after leaving the ground was given by h=-16t^2+32t.
:
How long he was in the air:
When he returns to the ground, h = 0 (also when he starts h = 0)
:
-16t^2 + 32t = 0
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Factor out -16t and you have:
-16t(t - 2) = 0
:
The two solutions:
-16t = 0
t = 0/-16
t = 0
and
t = +2 seconds, he was in the air
If you looked at on a graph it would resemble the flight path he takes:
t is on the x axis. Height is on the y axis
:
Notice that t = 0 when he starts and t = 2 when he returns to earth
:
:
For how long was he more than 14ft in the air?
The equation for that:
:
-16t^2 + 32t > 14
-16t^2 + 32t - 14 > 0
:
solve this using the quadratic formula: a=-16, b=32, c=-14
:
:
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Do the math here and you should get two solutions:
t > .6464 sec and t < 1.3536 sec
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We can say he exceeds 14 ft after .6464 sec & went below 14 ft at 1.3436 sec
The problems the problem wants to know how long he was above 14 ft
1.3436 - .6464 = .6972.
:
Let's say .7 seconds, that's roughly what the graph indicates also
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Did this make sense to you?
