Question

Equation of trajectory of a projectile is given by \(y=-x^{2}+10 x\) where \(x\) and \(y\) are in meters and \(x\) is along horizontal and \(y\) is verticall \(y\) upward and particle is projected from origin. Then which of the following options is incorrect: \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (1) initial velocity of particle is \(\sqrt{505} \mathrm{~m} / \mathrm{s}\) (2) horizontal range is \(10 \mathrm{~m}\) (3) maximum height is \(25 \mathrm{~m}\) (4) angle of projection with horizontal is \(\tan ^{-1}(5)\)

Short answer

All given options are correct; none is incorrect.

Step-by-step solution

Understand the Given Equation

The equation of the trajectory of the projectile is given by \[ y = -x^2 + 10x \] where \( y \) is the vertical position and \( x \) is the horizontal position.

Find the Horizontal Range

To find the horizontal range, set \( y = 0 \) (when the projectile hits the ground). \[ 0 = -x^2 + 10x \] Solve for \( x \). \[ x(x - 10) = 0 \] So, \( x = 0 \) or \( x = 10 \). Thus, the horizontal range is \( 10 \) meters.

Calculate the Maximum Height

To find the maximum height, use the vertex form of a parabola. The vertex form of \( y = ax^2 + bx + c \) gives the vertex at \( x = -\frac{b}{2a} \). Here, \( a = -1 \) and \( b = 10 \). \[ x = -\frac{10}{2(-1)} = 5 \] Substitute \( x = 5 \) into the equation to find \( y \). \[ y = -(5)^2 + 10(5) \] \[ y = -25 + 50 = 25 \] Thus, the maximum height is \( 25 \) meters.

Determine the Angle of Projection

The trajectory is described by a quadratic equation, hence the form is \( y = x \tan(\theta) - \frac{g x^2}{2u^2 \cos^2(\theta)} \), where \( \theta \) is the angle of projection and \( u \) is the initial velocity. In this case, comparing coefficients \( \tan(\theta) = 10 \) and \( \frac{g}{2u^2 \cos^2(\theta)} = 1 \). This leads to \( \theta = \tan^{-1}(5) \).

Find the Initial Velocity

Use the fact that \( \tan(\theta) = 5 \) and break it down: \( \cos(\theta) = \sqrt{1-\sin^2(\theta)} \). With \( \tan(\theta) = 5 \), then \( \sin(\theta) = \frac{5}{\sqrt{26}} \) and \( \cos(\theta) = \frac{1}{\sqrt{26}} \). Substituting these values back, \( \frac{g}{2u^2 \cos^2(\theta)} = 1 \) gives \( u^2 = \frac{10}{1/26} \). Thus, \( u = \sqrt{505} \) m/s.

Identify the Incorrect Statement

Based on the calculations: \( u = \sqrt{505} \) m/s is the initial velocity, horizontal range is 10m, and maximum height is 25m, and the angle of projection is \( \tan^{-1}(5) \). All given options are correct, thus none is incorrect.

Key concepts

Equation of Trajectory

Projectile motion can be exciting! One of the essential aspects involves understanding its equation of trajectory. In our problem, the trajectory equation is given as:
\[ y = -x^2 + 10x \] This equation describes the path that the projectile follows. Here, x and y are measured in meters. The x coordinate represents the horizontal distance, and y gives the vertical position.
The biggest takeaway? This quadratic equation implies the motion creates a parabolic path.

Horizontal Range

Next, let's talk about the horizontal range of the projectile. The horizontal range is the total distance a projectile travels along the x-axis before hitting the ground. In our scenario, we set y = 0 (meaning the projectile has landed).
Solving the equation:
\[ 0 = -x^2 + 10x \] gives us:
\[ x(x - 10) = 0 \] So, x can be either 0 or 10. Thus, the horizontal range is 10 meters.

• This is the distance the projectile covers horizontally before hitting the ground.

Maximum Height

Now, let's find out how high the projectile goes – also known as its maximum height. Maximum height is the peak vertical position the projectile reaches. To find this for our parabola y = -x^2 + 10x, we use the vertex formula:
\[ x = -\frac{b}{2a} \] Here, a = -1 and b = 10. So, the x value at the vertex will be:
\[ x = -\frac{10}{2(-1)} = 5 \] Substitute x = 5 back into the equation to find y:
\[ y = -(5)^2 + 10(5) \] \[ y = -25 + 50 = 25 \]
This makes the maximum height 25 meters.

Angle of Projection

To determine the angle at which the projectile was launched, we need the angle of projection. For our parabola, we know that the general formula for the trajectory is:
\[ y = x \tan(\theta) - \frac{g x^2}{2u^2 \cos^2(\theta)} \]
Comparing this to our equation y = -x^2 + 10x, we find that \tan(\theta) is 10.
This means:
\[ \theta = \tan^{-1}(5) \]
Thus, the angle of projection with the horizontal is \( \tan^{-1}(5) \).

Initial Velocity

Lastly, let's determine the initial velocity of our projectile.
We know:
\( \tan(\theta) = 5 \), and from trigonometric identities, this gives:
\[ \sin(\theta) = \frac{5}{\sqrt{26}} \] \[ \cos(\theta) = \frac{1}{\sqrt{26}} \]
By using these values in the formula \( \frac{g}{2u^2 \cos^2(\theta)} = 1 \), we get:
\[ u^2 = \frac{10}{1/26} = 505 \]
Therefore, the initial velocity \( u \) is \( \sqrt{505} \) m/s.