Mechanics- general motion in a straight line.

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SUMMARY

The discussion centers on the mechanics of a ball bearing fired vertically upwards through a tub of butter, with its upward velocity described by the equation v=13-10t-3t² cm/s. The participants calculated the time when the ball bearing comes to rest and the distance traveled upwards, resulting in q(a)=1s and q(b)=7cm. The challenge lies in solving for the time taken for the ball bearing to fall back to its original position after momentarily resting, using the equation ∫₀ᵗ 10T dT = 7.

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I calculated q(a)=1s
q(b)=7cm
I don't understand q(c)
 
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Fix the image to make it readable or type out the problem statement yourself.
 
skeeter said:
Fix the image to make it readable or type out the problem statement yourself.
Apologies.
A ball bearing is fired vertically upwards in a straight line through a tub of butter. The upward velocity of the ball bearing is given by v=13-10t-3t^2 cm/s, where t is the time from when it was fired upwards.
a) Find the time when the ball bearing comes momentarily to rest
b) Find how far the ball bearing has traveled upwards at this time

The ball bearing then falls downwards through the hole it has made in the butter. The downward velocity of the ball bearing is given by v= 10T cm/s, where T is the time from when it was momentarily at rest.
C) Find the time that the ball bearing takes ( from when it was momentarily at rest) to fall to its original position.
Iam not understand how to calculate (c)
 
$\displaystyle \int_0^t 10T \, dT = 7$

solve for $t$
 
skeeter said:
$\displaystyle \int_0^t 10T \, dT = 7$

solve for $t$
Thank you very much!
 

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