Question

List of Equations (Use only if the problem must be solved in terms of vectors) 3) d,-xf-Xi or 4) d=d1 + d2 + dy-yr-yi (Do not use if #2 applies) (Use only if the problem must be solved in terms of vectors) 7) uk dx/ t or vy dy/ t (Do not use if #6 applies) ,8) s=l/ht 9) [Cycle Length]/[Cycle Time] 10) [Cycle Length] x [Cycle Rate] 11) V2 = V1 + U2/1 (Do not use if #7 or #9 applies) 13) (vr-vi)/dt Δν/Δε (Use only if the problem must be solved in terms a or of vectors) (Do not use if #13 applies) (Do not use if #7 applies) (Do not use if #7 applies) 泌, 16) に(v, + ur)at d-v, t + ()a(At)2 Note: In any equation that includes At, you may substitute (ty -t) in place of At and consider it to be the same equation.

Answer 1

I assume that u need to solve Question 13.

13i) since the time of flight is 0.71s, and the horizontal velocity is constant( as there is no force acting on the object, other than gravity, which acts on the vertical velocity alone) we can use the equation of $d_{x}=v_{x}*\Delta t$

here $v_{x}=8.2*cos(20)$

13 ii) a) False.. this is because as the projection velocity decreases, the vertical component of the projection velocity also decreases, which would mean that the equivalent distance covered in the same time would be less.

13 ii) b)False.... The flight distance is maximum at 45 degrees and it decreases as we move away from 45 in both directions (that is either to 0 degrees or 90 degrees). so when the angle is decreased to 18 from 20 degrees, the flight distance also decreases.

13 ii)c) this would mean that the person landed on a platform, which is lower than the height from which he was projected. so this would essentially mean that he covered extra distance than what he would if he had landed at the same height. so the answer is true