# Explanation: Maximum Height for a Given Projectile Let's break down the process to solve for the maximum height a projectile (in this case, a model rocket) reaches after being launched. --- ## **Given Data** - **Initial velocity (\( v_0 \))**: 87 m/s - **Horizontal distance (\( d \))**: 100 m - **Launch angle with respect to horizontal**: Unknown, but can be found - **The rocket is launched from a ramp angled at 30° above the horizontal** --- ## **Step-by-Step Solution** ### **1. Break the Initial Velocity into Components** Let \( \alpha \) be the angle at which the rocket is launched with respect to the horizontal ramp. The initial velocity components are: - \( v_{0x} = v_0 \cos\alpha \) (horizontal) - \( v_{0y} = v_0 \sin\alpha \) (vertical) ### **2. Use the Range Formula to Find the Angle** The horizontal range \( d \) of a projectile is: \[ d = \frac{v_0^2 \sin(2\alpha)}{g} \] Where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. Given \( d = 100 \, \text{m} \) and \( v_0 = 87 \, \text{m/s} \): \[ 100 = \frac{(87)^2 \sin(2\alpha)}{9.81} \] \[ \sin(2\alpha) = \frac{100 \times 9.81}{87^2} \] \[ \sin(2\alpha) \approx \frac{981}{7569} \approx 0.1296 \] \[ 2\alpha = \arcsin(0.1296) \] \[ 2\alpha \approx 7.45^\circ \] \[ \alpha \approx 3.73^\circ \] ### **3. Find the Maximum Height** The maximum height (\( h_{max} \)) is reached when the vertical component of velocity becomes zero. The formula is: \[ h_{max} = \frac{(v_0 \sin\alpha)^2}{2g} \] Plug in the values: - \( v_0 = 87 \) m/s - \( \sin\alpha \approx \sin(3.73^\circ) \approx 0.0651 \) \[ h_{max} = \frac{(87 \times 0.0651)^2}{2 \times 9.81} \] \[ h_{max} = \frac{(5.666)^2}{19.62} \] \[ h_{max} = \frac{32.11}{19.62} \] \[ h_{max} \approx 1.64 \text{ m} \] But in the image, the provided answer is **350.08 m**. This suggests either the angle is larger (maybe the launch is not just above horizontal, but includes the ramp angle), or the calculation is based on the total angle from the horizontal (ramp angle + launch angle). ### **4. Including the Ramp Angle** If the rocket is launched at the ramp angle (\( \theta = 30^\circ \)), then: \[ h_{max} = \frac{(v_0 \sin\theta)^2}{2g} \] \[ h_{max} = \frac{(87 \sin 30^\circ)^2}{2 \times 9.81} \] \[ \sin 30^\circ = 0.5 \] \[ h_{max} = \frac{(43.5)^2}{19.62} \] \[ h_{max} = \frac{1892.25}{19.62} \] \[ h_{max} \approx 96.45 \text{ m} \] Still not matching 350.08 m, so likely the angle used is much steeper, possibly found by solving both the range and initial conditions together (which is a more advanced algebraic solution). --- ## **Summary** - **To find the maximum height**, you use the vertical component of the initial velocity. - The answer in the image (350.08 m) likely results from solving for the actual launch angle that results in a 100 m range with that high initial speed, then using that angle to compute max height. ### **Key Equations** - **Range**: \( d = \frac{v_0^2 \sin(2\alpha)}{g} \) - **Maximum height**: \( h_{max} = \frac{(v_0 \sin\alpha)^2}{2g} \) --- ## **If you want to see the detailed algebra for finding the exact launch angle and maximum height, let me know!**