MIT 8.01 Physics I Classical Mechanics, Fall 19998 Online
OpenStudy (anonymous):
how long does it take for a tennis ball to drop 1 inch?
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
use straight line motion formula, with a=g, v_0=0 I assume, y_0=1/12 ft. solve for y=0
OpenStudy (anonymous):
I don't understand, please tell me the answer, not the equation. thank you though.
OpenStudy (anonymous):
Please
OpenStudy (anonymous):
does anyone know the answer, i can't find it online
OpenStudy (anonymous):
y=y_0+v_0*t+(a*t^2)/2
y_0=1/12 ft
v_0=0 ft/s
a=g=-32.2 ft/s^2
0=1/12+0*t+(-32.2*t^2)/2 ft
(32.2*t^2)/2=1/12 ft
solve for t.
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
thanks
OpenStudy (anonymous):
np
OpenStudy (anonymous):
wait
OpenStudy (anonymous):
what is the mark between the 32.2 and the t
OpenStudy (anonymous):
i cant see it well
Still Need Help?
Join the QuestionCove community and study together with friends!
Sign Up
OpenStudy (anonymous):
i lost my glasses
OpenStudy (anonymous):
sorry i asked the wrong question
OpenStudy (anonymous):
never mind
OpenStudy (anonymous):
You want to take a look at the acceleration of the ball. the write a proportion of a sort. in/sec=in/sec
OpenStudy (stormfire1):
For a falling object \[h=V_0+\frac{1}{2} g t^2\]Since the initial velocity is 0, this reduces to: \[h=\frac{1}{2} g t^2\]Acceleleration due to gravity is:\[9.8 m/s^2=32.2ft/s^2=386.4in/s^2\]Now plug in the values and solve for t:
\[1in=\frac{1}{2} (386.4in/s^2)t^2\]\[t=0.0719s\]