Benjamin V.

asked • 04/08/17

Space 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.
 

Arturo O.

What is your question?  By the way, if the upward direction is positive, the term with g needs a minus sign:
 
h(t) = -(1/2)gt2 + v0t + h0
 
Also, g is the acceleration of gravity.  The force of gravity (i.e. the weight) is mg, where m is the mass of the ball.
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04/08/17

Benjamin V.

There is no negative sign so I'm party sure it's a downward parabola, the question reads 
Part 1 Frank Mason III is 6 feet tall, so we can estimate that he releases the ball from 7 ft.
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04/08/17

Benjamin V.

In the air when he is shooting. The average initial velocity of his shots has been computed at 24ft/sec. Find the force of gravity on earth in ft/sec2 . Then, using these numbers, find the height of a basketball shot by MASON with respect to time and anwser the following questions
1. At what time does the ball reach 10 feet? Give both the exact anwser(s) and the rounded anwser(s) to the nearest hundredth 
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04/08/17

Andrew M.

The parabolic flight of the basketball will be a parabola
opening downward.  The equation is:
 
h(t) = (-1/2)gt2 + v0t + h0
g = 32 ft/sec2, v0 = 24 ft/sec, h0 = 7 ft
 
h(t) = -16t2 + 24t + 7
 
For height of 10 ft:
-16t2 + 24t + 7 = 10
-16t2 + 24t - 3 = 0
Using quadratic equation:
 
t = [-24±√(242-4(-16)(-3))]/(2(-16))
t = (-24±√384)/(-32)
t = (-24± 19.59592)/(-32)
 
Note:  the basketball will pass through 10 ft
twice... On the way up when first thrown, and
on the way back down.
 
t = 0.14 seconds and t = 1.36 seconds
 
 
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04/08/17

1 Expert Answer

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Andrew M. answered • 04/08/17

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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?
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Andrew M.

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

Arturo O.

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

Andrew M.

No problem Benjamin.  You're very welcome.
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04/09/17

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