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 $9.8 m/sec^2$.
