Answer
(a) The speed at the top is 2.3 m/s
(b) The ice cube's speed at an angle of $\theta$ is $\sqrt{(7.04~m^2/s^2)+(1.96~m^2/s^2)~cos(\theta)}$
Work Step by Step
(a) We can use conservation of energy to solve this question. Note that the height at the top of the pipe is 20 cm.
$KE_2+PE_2 = KE_1+PE_1$
$\frac{1}{2}mv_2^2+mgh=\frac{1}{2}mv_1^2+0$
$v_2^2=v_1^2-2gh$
$v_2= \sqrt{v_1^2-2gh}$
$v_2= \sqrt{(3.0~m/s)^2-(2)(9.80~m/s^2)(0.20~m)}$
$v_2 = 2.3~m/s$
The speed at the top is 2.3 m/s
(b) We can find the height $h$ above the bottom of the pipe when the angle is $\theta$. Let $R$ be the radius of the pipe.
$\frac{R-h}{R} = cos(\theta)$
$h = R~(1-cos(\theta))$
We can use conservation of energy to find the speed when the ice cube is at an angle $\theta$. Note in part (a) that: $v_2= \sqrt{v_1^2-2gh}$
$v_2= \sqrt{v_1^2-2gR~(1-cos(\theta))}$
$v_2= \sqrt{v_1^2- 2gR+2gR~cos(\theta)}$
$v_2= \sqrt{(3.0~m/s)^2-(2)(9.80~m/s^2)(0.10~m)+(2)(9.80~m/s^2)(0.10~m)~cos(\theta)}$
$v_2= \sqrt{(7.04~m^2/s^2)+(1.96~m^2/s^2)~cos(\theta)}$
The ice cube's speed at an angle of $\theta$ is $\sqrt{(7.04~m^2/s^2)+(1.96~m^2/s^2)~cos(\theta)}$
