Answer
a) $t=0.405s$
b) $t=1.017s$
c) the bale's time in the air increases.
Work Step by Step
(a) We know that
$5m/s=\sqrt{(1.12m/s)^2+v_y^2}$
This simplifies to:
$v_y=\pm 4.873m/s$
As the bale is rising, so we take the positive value of $v_y$.
Now $v_y=v_{\circ y}-gt$
This can be rearranged as:
$t=\frac{v_{\circ y}-v_y}{g}$
We plug in the known values to obtain:
$t=\frac{8.85m/s-4.873m/s}{9.8m/s^2}$
$t=0.405s$
(b) We know that
$t=\frac{v_{\circ y}-v_y}{g}$
We plug in the known values to obtain:
$t=\frac{8.85m/s-(-1.12m/s)}{9.8m/s}$
$t=1.017s$
(c) We know that when the bale is tossed straight up with the same initial speed, then the vertical component of the initial velocity is greater and the bale rises higher in the air. So, we conclude that the bale's time in the air increases.
