Question

the turd is launched at a speed of 50 m/s at an angle of 62 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 velocity

The initial vertical velocity \( v_{0y} = v_0 \sin\theta \), where \( v_0 = 50\,\text{m/s} \) and \( \theta = 62^\circ \). So \( v_{0y} = 50 \sin(62^\circ) \approx 50 \times 0.8829 = 44.145\,\text{m/s} \).

Step2: Use kinematic equation for maximum height

At maximum height, vertical velocity \( v_y = 0 \). The kinematic equation \( v_y^2 = v_{0y}^2 - 2gh \) (where \( g = 9.8\,\text{m/s}^2 \)) gives \( h=\frac{v_{0y}^2}{2g} \). Substituting values: \( h=\frac{(44.145)^2}{2\times9.8} \approx \frac{1948.78}{19.6} \approx 99.43\,\text{m} \).

Answer:

Approximately \( 99.4\,\text{m} \) (or more precisely \( \approx 99.43\,\text{m} \))