A projectile is thrown from origin in vertical plane (x along horizontal, +y in vertical upwards) with speed of 10 m/s that passes through a point P(6, 3), what may be the angle of... A projectile is thrown from origin in vertical plane (x along horizontal, +y in vertical upwards) with speed of 10 m/s that passes through a point P(6, 3), what may be the angle of projection with horizontal for the projectile? (take g = 10 m/s²)
Understand the Problem
The question asks for the angle of projection of a projectile thrown from the origin that passes through a specific point. It provides the speed of the projectile and the acceleration due to gravity to assist in calculating the angle.
Answer
The angle of projection is $37^\circ$.
Answer for screen readers
The angle of projection is approximately $37^\circ$.
Steps to Solve
- Identify the problem parameters
We have a projectile launched from the origin with an initial speed of $10 , \text{m/s}$, passing through the point $P(6, 3)$. The acceleration due to gravity $g$ is given as $10 , \text{m/s}^2$.
- Break down the calculations
Using the equations of projectile motion, the horizontal distance $R$ is given by:
$$ R = \frac{v^2 \sin(2\theta)}{g} $$
where $R$ is the horizontal range, $v$ is the initial speed, and $\theta$ is the angle of projection.
Here, $R = 6 , \text{m}$ and $v = 10 , \text{m/s}$.
- Calculate the vertical distance time at point P
The vertical distance $y$ at any time $t$ is given by:
$$ y = v \sin(\theta) t - \frac{1}{2} g t^2 $$
In this case, we also know from the equations of motion that:
$$ t = \frac{R}{v \cos(\theta)} = \frac{6}{10 \cos(\theta)} $$
- Substitute $t$ into the vertical distance equation
Substituting $t$ into the vertical distance equation gives:
$$ 3 = 10 \sin(\theta) \left(\frac{6}{10 \cos(\theta)}\right) - \frac{1}{2} g \left(\frac{6}{10 \cos(\theta)}\right)^2 $$
Simplifying this:
$$ 3 = \frac{6 \sin(\theta)}{\cos(\theta)} - \frac{1}{2} g \frac{36}{100 \cos^2(\theta)} $$
- Substituting the value of g
Substituting $g = 10 , \text{m/s}^2$ into the simplified equation yields:
$$ 3 = 6 \tan(\theta) - \frac{18}{10 \cos^2(\theta)} $$
- Rearranging the equation
Bringing all terms to one side gives us:
$$ 6 \tan(\theta) - \frac{18}{10 \cos^2(\theta)} - 3 = 0 $$
This can be solved for $\theta$ using numerical methods or estimation.
- Calculate the angle
Testing the options A, B, C, D through calculations or using a calculator:
By trial, we find the angle $\theta$ that gets closest to fulfilling this equation is approximately $36.87^\circ \approx 37^\circ$.
The angle of projection is approximately $37^\circ$.
More Information
This calculation utilizes principles of projectile motion, particularly the relationship between the horizontal range, vertical height, and angle of projection.
Tips
- Misinterpreting the angles—it's important to use the correct trigonometric identities to relate the angles in the calculations.
- Neglecting the role of gravity while substituting values, which can lead to incorrect results.
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