d = vi * t + ½ * a * t^2
d = Final height = 60 * sin 10
vi = Initial vertical velocity = v * 0.5
a = -9.8 m/s^2
60 * sin 10 = v * 0.5 * t + ½ * -9.8 * t^2
60 * sin 10 = v * 0.5 * t – 4.9 * t^2
Substitute (60 * cos 10˚ ÷ v * cos 30˚) for t
60 * sin 10 = v * 0.5 * (60 * cos 10˚ ÷ v * cos 30˚) – 4.9 * (60 * cos 10˚ ÷ (v * cos 30˚)^2
v * 0.5 * (60 * cos 10˚ ÷ v * cos 30˚) = 30 * cos 10 ÷ cos 30
(60 * cos 10˚ ÷ v * cos 30˚)^2 = 3600 * cos^2 10˚ ÷ (v^2 * 0.75)
3600 * cos^2 10˚ ÷ (v^2 * 0.75) = 4800 * cos^2 10˚ ÷ v^2
4.9 * 4800 * cos^2 10˚ ÷ v^2 = 23,520 * cos^2 10˚ ÷ v^2
60 * sin 10 = (30 * cos 10 ÷ cos 30) – 23,520 * cos^2 10˚ ÷ v^2
60 * sin 10 – (30 * cos 10 ÷ cos 30) = -23,520 * cos^2 10˚ ÷ v^2
Multiply both sides by v^2
v^2 * [60 * sin 10 – (30 * cos 10 ÷ cos 30)] = -23520 * cos^2 10˚
v^2 = (-23520 * cos^2 10˚) ÷ [60 * sin 10 – (30 * cos 10 ÷ cos 30)]
v^2 = -22,810.78522 ÷ -23.69585062
v = √962.6490682
This is approximately 31 m/s.
Check:
t = 60 * cos 10 ÷ 31 * cos 30
This is approximately 2.2 s
60 * sin 10 = 31* 0.5 * 2.2 – 4.9 * 2.2^2
10.41889066 = 10.384
This proves that the initial velocity is approximately 31 m/s.