Question

the turd is launched at a speed of 44 m/s at an angle of 51 degrees above the horizontal. at what height should a portal be placed in order to intercept the turd at its highest point above the ground?

Explanation:

Step1: Find vertical component of initial velocity

The vertical component \( v_{0y} = v_0 \sin\theta \), where \( v_0 = 44 \, \text{m/s} \), \( \theta = 51^\circ \).
\( v_{0y} = 44 \sin(51^\circ) \approx 44 \times 0.7771 \approx 34.19 \, \text{m/s} \)

Step2: Use kinematic equation for maximum height

At maximum height, vertical velocity \( v_y = 0 \). The equation \( v_y^2 = v_{0y}^2 - 2gh \) (where \( g = 9.8 \, \text{m/s}^2 \)) gives \( h = \frac{v_{0y}^2}{2g} \).
\( h = \frac{(34.19)^2}{2 \times 9.8} \approx \frac{1169.0}{19.6} \approx 59.64 \, \text{m} \)

Answer:

Approximately \( 59.6 \, \text{m} \) (or more precisely \( \approx 60 \, \text{m} \) depending on precision)