A stone is projected from the ground with a velocity of 14 m/s, one second later it clears a wall 2 m high. The angle of projection is?
Understand the Problem
The question asks for the angle of projection of a stone that is thrown from the ground with a specified velocity and clears a wall of a certain height after one second. The goal is to determine the correct angle from the provided options.
Answer
The angle of projection is $30^\circ$.
Answer for screen readers
The angle of projection is $30^\circ$.
Steps to Solve
- Identify Initial Parameters
Given parameters:
- Initial velocity ($u$) = 14 m/s
- Time ($t$) = 1 s
- Height of the wall ($h$) = 2 m
- Acceleration due to gravity ($g$) = 10 m/s²
- Break Down the Velocity
The initial velocity can be broken down into horizontal and vertical components:
- Vertical component: $u_y = u \cdot \sin(\theta)$
- Horizontal component: $u_x = u \cdot \cos(\theta)$
- Calculate the Vertical Displacement
The vertical displacement after 1 second: $$ h = u_y \cdot t - \frac{1}{2} g t^2 $$
Substituting known values: $$ 2 = (u \cdot \sin(\theta) \cdot 1) - \frac{1}{2} \cdot 10 \cdot 1^2 $$
- Simplify the Equation
We can rewrite the equation: $$ 2 = 14 \cdot \sin(\theta) - 5 $$ Adding 5 to both sides gives: $$ 7 = 14 \cdot \sin(\theta) $$
- Solve for $\sin(\theta)$
Dividing both sides by 14: $$ \sin(\theta) = \frac{7}{14} = \frac{1}{2} $$
- Determine the Angle
To find the angle $\theta$: $$ \theta = \arcsin\left(\frac{1}{2}\right) $$ This results in: $$ \theta = 30^\circ $$
The angle of projection is $30^\circ$.
More Information
In projectile motion, the angle of launch significantly affects the maximum height and distance traveled. Here, $\theta = 30^\circ$ produces the right vertical displacement to clear a 2 m high wall after one second.
Tips
- Miscalculating the vertical displacement or forgetting to account for gravity in the equation could lead to an incorrect angle.
- Assuming the height reached is simply the vertical component multiplied by time without considering gravitational effects.
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