Question 11.191: Sand is discharged at A from a conveyor belt and falls onto ...

The motion is projectile motion. Place the origin at Point A. Then x_{0}=0 and y_{0}=0.

The coordinates of Point B are x_{B}=30 \mathrm{\ ft} and y_{B}=-18 \mathrm{\ ft}.

Horizontal motion:

\begin{aligned}\nu_{x} & =\nu_{0} \cos 25^{\circ} & \text{(1)} \\x & =\nu_{0} t \cos 25^{\circ} & \text{(2)}\end{aligned}

Vertical motion:

\begin{aligned}\nu_{y} & =\nu_{0} \sin 25^{\circ}-g t & \text{(3)} \\y & =\nu_{0} t \sin 25^{\circ}-\frac{1}{2} g t^{2} & \text{(4)}\end{aligned}

At Point B, Eq. (2) gives

\nu_{0} t_{B}=\frac{x_{B}}{\cos 25^{\circ}}=\frac{30}{\cos 25^{\circ}}=33.101 \mathrm{\ ft}

Substituting into Eq. (4),

\begin{aligned}-18 & =(33.101)\left(\sin 25^{\circ}\right)-\frac{1}{2}(32.2) t_{B}^{2} \\t_{B} & =1.40958 \mathrm{~s}\end{aligned}

(a) Speed of the belt.

\nu_{0}=\frac{\nu_{0} t_{B}}{t_{B}}=\frac{33.101}{1.40958}=23.483

\nu_{0}=23.4 \mathrm{\ ft} / \mathrm{s}\blacktriangleleft

Eqs. (1) and (3) give

Components of acceleration.

(b) Radius of curvature at B.

\begin{aligned}& a_{n}=\frac{\nu^{2}}{\rho} \\& \rho=\frac{\nu^{2}}{a_{n}}=\frac{(41.36)^{2}}{16.57} & \rho=103.2 \mathrm{\ ft}\blacktriangleleft\end{aligned}