## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.4 - The Derivative as a Rate of Change - Exercises - Page 144: 14

#### Answer

a) The equation for the ball's velocity during free fall is $v=9.8t(m/sec)$. b) A freely falling body near the surface of the Earth experiences a constant acceleration of $9.8 m/sec^2$.

#### Work Step by Step

$$v=9.8(\sin\theta)t(m/sec)$$ a) During free fall, the angle $\theta$ that the ball makes with the ground equals $90^\circ$, so $\sin\theta=\sin90^\circ=1$. Therefore, the equation for the ball's velocity during free fall is $$v=9.8\times1\times t=9.8t(m/sec)$$ b) We need to find the acceleration $a$ according to the equation for $v$ found in part a). $$a=\frac{dv}{dt}=\frac{d}{dt}(9.8t)=9.8m/sec^2$$ Therefore, a freely falling body near the surface of the Earth experiences a constant acceleration of $9.8 m/sec^2$.

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