Using
We get R = 19.76 m
time of flight =
= 2.756s
We can calculate maximum height reached by projectile above 9m scene using
h = v^{2}sin^{2}/2g = 5.35 m
So maximum height above ground = 9+5.35 = 14.35m
Speed along vertical = v_{initial -}gt = 12.5sin 55 - 9.8*2.756 = -16.77m/s
Speed along horizontal = vcos = 12.5cos55 = 7.17 m/s
so v_{net} = 18.24m/s
For angle =tan^{-1}(16.77/7.17) = 66.85^{o} below horizontal
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