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# An object is projected upward with initiall velocity $v_0$ meters per second from a point $s_0$ meters above the ground. Show that$$[v(t)]^2 = v_0^2 - 19.6[s(t) - s_0]$$

## $$[v(t)]^{2}=v_{0}^{2}-19.6\left[s(t)-s_{0}\right]$$

Derivatives

Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

recall the fact that acceleration a of t is negative. 9.8 in the units are meters per second squared. So we know that velocity is the integral of accelerations of the integral of negative 9.8 d t. Thus, if we integrate, we know the equation would be negative 9.8 times the variable of tea, plus our constant of C. Therefore, we know he can write this as position being the integral of velocity and we can write via, oh, as our constant. You know, its initial velocity. This is negative for 0.9 cheese square plus v a vote times are very bulls t plus c. Then we know, but we can write V F G squared is equivalent to vivo squared. Remember, this is initial velocity squared minus 99 point in terms to 19 point sex. And then this is Asif teed position minus initial position. That's us of O

#### Topics

Derivatives

Differentiation

Volume

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp