Problem 7.20

กำหนดให้ \(u = 22.4 m/s, y = -102 m, θ = 37.0°\) และ \(g = 9.80 m/s^{2}\)

a) จาก \(y = u_{y}t+\frac{1}{2}a_{y}t^{2}\)

\(y=utsin\theta +\frac{1}{2}(-g)t^{2}\)

\(4.90t^{2} - 13.48t - 102 = 0\)

จะได้ \(t = -3.39 s, 6.14 s \to \therefore t = 6.14 s\)

จาก \(x = u_{x}t = utcos\theta = 110 m\)

จาก \(r = \sqrt{x^{2} + y^{2}}=\sqrt{(110)^{2} + (-102)^{2}}= 150 m \;\blacksquare\)

b) จาก \(v^{2} = u^{2} + 2\vec{a}·\vec{s}\)

ĵîĵ

\(v^{2} = (22.4)^{2} + 2(-9.80 )·(110 - 102 )\)

\(v = 50.0 m/s \;\blacksquare\)