✔ 最佳答案
(a) (i) x = V*cosΘ*t = 20m/s * cos30º * 1.2s = 21 m
y = V*sinΘ*t + ½at² = 20m/s*sin30º*1.2s - ½*9.8m/s²*(1.2s)²
y = 4.9 m
ii) Vx = 20m/s*cos30º = 17 m/s
Vy = 20m/s*sin30º - 9.8m/s²*1.2s = -1.8 m/s
|V| = √(Vx² + Vy²) = 17.4 m/s ≈ 17 m/s
Θ = arctan(Vy/Vx) = -5.8º = 5.8º below horizontal
(b) Assuming the ball is launched from ground level,
y = v*sinΘ*t + ½at²
so here 4 = 10t - 4.9t²
quadratic with solutions at t = 0.55 s and t = 1.5 s
(c) Still 4 = 10t - 4.9t²
and so the answers are the same
Hope this helps!