Question

the turd is launched at a speed of 46 m/s at an angle of 32 degrees above the horizontal. at exactly what time after launch should the ground be covered in a portal to intercept the turd before it hits the ground?

Explanation:

Step1: Find vertical component of velocity

The initial vertical velocity \( v_{0y} = v_0 \sin\theta \), where \( v_0 = 46 \, \text{m/s} \) and \( \theta = 32^\circ \). So \( v_{0y} = 46 \sin(32^\circ) \approx 46 \times 0.5299 \approx 24.3754 \, \text{m/s} \).

Step2: Use vertical motion equation for time of flight

The vertical displacement \( y = v_{0y}t - \frac{1}{2}gt^2 \). When it hits the ground, \( y = 0 \) (assuming launch from ground level). So \( 0 = 24.3754t - 4.9t^2 \) (since \( g = 9.8 \, \text{m/s}^2 \), \( \frac{1}{2}g = 4.9 \)). Factor out \( t \): \( t(24.3754 - 4.9t) = 0 \). Solutions are \( t = 0 \) (launch time) and \( t = \frac{24.3754}{4.9} \approx 4.9746 \, \text{s} \).

Answer:

Approximately \( 5.0 \, \text{s} \) (or more precisely \( \approx 4.97 \, \text{s} \))