Benjamin V.
asked 04/08/17Space jammin' with parabolas!
h (t)= (1/2)gt2+vot+ho where he is the height of the ball in feet at time t in seconds, g is the force of gravity in feet per second squared, vo Is the initial velocity of feet per second, and ho is the initial height from here the ball is thrown.
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1 Expert Answer
Inactive Tutor answered 04/08/17
Tutor
New to Wyzant
Your equation is missing a negative sign:
h(t) = (-1/2)gt2 + v0t + h0
Also, g - the acceleration of gravity - can be
either 32 ft/sec2 or 9.8 m/sec2 depending on
whether you are working with feet or meters.
Did you have a question?
Inactive Tutor
The basketball will pass through 10 feet height twice:
Once on the way up when first thrown, and once more
on the way back down.
h(t) = -16t2 + 24t + 7
-16t2 + 24t + 7 = 10
-16t2 + 24t - 3 = 0
From quadratic equation:
t = [-24±√((24)2-4(-16)(-3))]/(2(-16))
t = [-24±√384]/(-32)
Finishing this you will get the two times
at which the basketball passes through
a height of 10 feet.
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04/09/17
Inactive Tutor
Benjamin,
Andrew gave you a correct solution, but note his h(t) equation has a negative sign in front of g, as it should, because the ball is always accelerating downward regardless of whether it is traveling up or down. And again, it is not correct to call g the force of gravity.
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04/09/17
Benjamin V.
You are seriously amazing Andrew! Thank you so much! I might need some help on the rest of the problems, just a heads up! :)
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04/09/17
Inactive Tutor
No problem Benjamin. You're very welcome.
Report
04/09/17
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Inactive Tutor
04/08/17