Question

the turd is launched at a speed of 46 m/s at an angle of 59 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 vertical component of the initial velocity \( v_{0y} \) is given by \( v_{0y} = v_0 \sin(\theta) \), where \( v_0 = 46 \, \text{m/s} \) and \( \theta = 59^\circ \). So, \( v_{0y} = 46 \sin(59^\circ) \). Calculating \( \sin(59^\circ) \approx 0.8572 \), we get \( v_{0y} \approx 46 \times 0.8572 \approx 39.43 \, \text{m/s} \).

Step2: Use vertical motion equation

The vertical displacement \( y \) as a function of time \( t \) is \( y = v_{0y}t - \frac{1}{2}gt^2 \), where \( g = 9.8 \, \text{m/s}^2 \). When the object hits the ground, \( y = 0 \). So, \( 0 = v_{0y}t - \frac{1}{2}gt^2 \). Factoring out \( t \), we get \( t(v_{0y} - \frac{1}{2}gt) = 0 \). One solution is \( t = 0 \) (launch time), the other is \( v_{0y} - \frac{1}{2}gt = 0 \). Solving for \( t \): \( t = \frac{2v_{0y}}{g} \).

Step3: Calculate time

Substitute \( v_{0y} \approx 39.43 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \) into the formula: \( t = \frac{2 \times 39.43}{9.8} \approx \frac{78.86}{9.8} \approx 8.05 \, \text{s} \).

Answer:

Approximately \( 8.05 \, \text{seconds} \)