Question

the turd is launched at a speed of 46 m/s at an angle of 49 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 initial vertical velocity \( v_{0y} = v_0 \sin\theta \), where \( v_0 = 46 \, \text{m/s} \) and \( \theta = 49^\circ \). So \( v_{0y} = 46 \sin(49^\circ) \approx 46\times0.7547 \approx 34.72 \, \text{m/s} \).

Step2: Find time of flight

Using the vertical motion equation \( v_y = v_{0y}-gt \), at maximum height \( v_y = 0 \), but for total time (time to hit ground, assuming launch and landing at same height), we can use \( y = v_{0y}t-\frac{1}{2}gt^2 \), and when \( y = 0 \) (ground level), \( 0 = v_{0y}t-\frac{1}{2}gt^2 \). Factoring out \( t \), \( t(v_{0y}-\frac{1}{2}gt)=0 \). So \( t = 0 \) (launch) or \( t=\frac{2v_{0y}}{g} \). Substituting \( v_{0y}\approx34.72 \, \text{m/s} \) and \( g = 9.8 \, \text{m/s}^2 \), \( t=\frac{2\times34.72}{9.8}\approx\frac{69.44}{9.8}\approx7.09 \, \text{s} \).

Step3: Find horizontal component of velocity

The horizontal velocity \( v_{0x}=v_0\cos\theta \), so \( v_{0x}=46\cos(49^\circ)\approx46\times0.6561\approx30.18 \, \text{m/s} \).

Step4: Find horizontal range

Horizontal range \( R = v_{0x}t \). Substituting \( v_{0x}\approx30.18 \, \text{m/s} \) and \( t\approx7.09 \, \text{s} \), \( R\approx30.18\times7.09\approx214 \, \text{m} \) (approximate value, more precise calculation: \( \cos(49^\circ)\approx0.656059, \sin(49^\circ)\approx0.75470958 \), \( t=\frac{2\times46\times0.75470958}{9.8}\approx\frac{69.43327}{9.8}\approx7.085 \, \text{s} \), \( R = 46\times0.656059\times7.085\approx46\times4.65\approx213.9 \approx 214 \, \text{m} \))

Answer:

Approximately \( \boldsymbol{214 \, \text{meters}} \) (or more precisely, around 214 m depending on calculation precision)