Question

the turd is launched at a speed of 49 m/s at an angle of 64 degrees above the horizontal. how far horizontally from the catapult should a portal be placed in order to intercept the turd before it hits the ground?

Explanation:

Step1: Find vertical component of velocity

The vertical component of the initial velocity \( v_{0y} = v_0 \sin\theta \), where \( v_0 = 49 \, \text{m/s} \) and \( \theta = 64^\circ \). So \( v_{0y}=49\sin(64^\circ) \approx 49\times0.8988\approx44.04\,\text{m/s} \).

Step2: Find time of flight

Using the vertical motion equation \( y = v_{0y}t-\frac{1}{2}gt^2 \), when the object hits the ground \( y = 0 \). So \( 0 = 44.04t - 4.9t^2 \). Factoring out \( t \), we get \( t(44.04 - 4.9t)=0 \). The non - zero solution is \( t=\frac{44.04}{4.9}\approx8.99\,\text{s} \).

Step3: Find horizontal component of velocity

The horizontal component of the initial velocity \( v_{0x}=v_0\cos\theta \), so \( v_{0x}=49\cos(64^\circ)\approx49\times0.4384\approx21.48\,\text{m/s} \).

Step4: Find horizontal range

The horizontal range \( R = v_{0x}t \). Substituting \( v_{0x}\approx21.48\,\text{m/s} \) and \( t\approx8.99\,\text{s} \), we get \( R\approx21.48\times8.99\approx193.1\,\text{m} \).

Answer:

The horizontal distance is approximately \(\boldsymbol{193\,\text{m}}\) (or more precisely around \(193.1\,\text{m}\))