#### Answer

(a) At t = 0:
$v = (2.0\hat{i}+4.0\hat{j})~m/s$
At t = 2 s:
$v = (2.0\hat{i})~m/s$
At t = 3 s:
$v = (2.0\hat{i}-2.0\hat{j})~m/s$
(b) The magnitude of g is $2.0~m/s^2$
(c) $\theta = 63.4^{\circ}$

#### Work Step by Step

(a) At t = 1 s:
$v = (2.0\hat{i}+2.0\hat{j})~m/s$
At t = 2 s:
$v = (2.0\hat{i})~m/s$
At t = 3 s:
$v = (2.0\hat{i}-2.0\hat{j})~m/s$
We can see that the vertical component of velocity decreases at a rate of $2.0~m/s^2$. Therefore, at t = 0,
$v = (2.0\hat{i}+4.0\hat{j})~m/s$
(b) $v_y = v_{y0}+gt$
$g = \frac{v_y-v_{y0}}{t}$
$g = \frac{2.0~m/s-4.0~m/s}{1.0~s}$
$g = -2.0~m/s^2$
The magnitude of g is $2.0~m/s^2$
(c) We can find the launch angle above the horizontal.
$tan(\theta) = \frac{v_y}{v_x}$
$tan(\theta) = \frac{4.0~m/s}{2.0~m/s}$
$\theta = arctan(\frac{4.0~m/s}{2.0~m/s})$
$\theta = 63.4^{\circ}$