Answer
(a) t = 0.682 s and t = 2.99 s
(b) At t = 0.682 s:
$v_x = 24.0~m/s$
$v_y = 11.3~m/s$
At t = 2.99 s:
$v_x = 24.0~m/s$
$v_y = -11.3~m/s$
(c) The magnitude of the velocity is 30.0 m/s
The direction is an angle of $36.9^{\circ}$ below the horizontal.
Work Step by Step
(a) $v_{0y} = v_0~sin(\theta) = (30.0~m/s)~sin(36.9^{\circ})$
$v_{0y} = 18.0~m/s$
We can find an expression for the height above the point where the ball left the bat.
$y = v_{0y}~t-\frac{1}{2}gt^2$
$10.0~m = (18.0~m/s)~t - (4.90~m/s^2)~t^2$
$(4.90~m/s^2)~t^2 - (18.0~m/s)~t +10.0~m = 0$
We can use the quadratic formula to find $t$.
$t = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{(18.0)\pm \sqrt{(-18.0)^2-(4)(4.90)(10.0)}}{(2)(4.90)}$
$t = 0.682~s$ and $t = 2.99~s$
(b) At t = 0.682 s:
$v_x = v_0~cos(\theta) = (30.0~m/s)~cos(36.9^{\circ})$
$v_x = 24.0~m/s$
$v_y = v_{0y} - gt = (18.0~m/s) - (9.80~m/s^2)(0.682~s)$
$v_y = 11.3~m/s$
At t = 2.99 s:
$v_x = v_0~cos(\theta) = (30.0~m/s)~cos(36.9^{\circ})$
$v_x = 24.0~m/s$
$v_y = v_{0y} - gt = (18.0~m/s) - (9.80~m/s^2)(2.99~s)$
$v_y = -11.3~m/s$
(c) When the ball returns to the level at which it left the bat, the magnitude of the velocity is the same as the initial velocity and the angle is below the horizontal instead of above the horizontal.
The magnitude of the velocity is 30.0 m/s
The direction is an angle of $36.9^{\circ}$ below the horizontal.
