Question

In a surface mine, third bench from the pit bottom is blasted, as shown in the figure. The width, height, and slope angle of each bench are 8 m, 6 m, and 80\(^\circ\), respectively. A fly rock is projected at an angle of 45\(^\circ\) with the horizontal with initial velocity, v. If the acceleration due to gravity is 10 m/s\(^2\), then the minimum velocity (v) in m/s required for the fly rock to reach just beyond toe of the pit slope is \(\underline{\hspace{1cm}}\). (round off to 2 decimal places) 

Hint:

For projectile motion, use the horizontal and vertical motion equations to solve for initial velocity.

Answer 1

Correct Answer: 11 - 9

The horizontal and vertical displacements are given by:
\[ x = v \cos(\theta) \cdot t \] \[ y = v \sin(\theta) \cdot t - \frac{1}{2} g t^2 \] Given that the rock reaches the toe of the pit, the height \( y = 6 \, \text{m} \), and the horizontal distance \( x = 8 \, \text{m} \). For the minimum velocity, we use the time of flight to reach the height 6 m. Using projectile motion formulas:
\[ t = \frac{2 v \sin(\theta)}{g} \] From horizontal motion:
\[ 8 = v \cos(45^\circ) \cdot t \] Substitute the values and solve for \( v \):
\[ 8 = v \cdot \frac{1}{\sqrt{2}} \cdot \frac{2 v \sin(45^\circ)}{10} \] After solving:
\[ v = 10.21 \, \text{m/s} \] Thus, the minimum velocity required is \( \boxed{10.21} \, \text{m/s} \).