Answer
a)13.6m
b)34.6m/s
c)103m
Work Step by Step
\ltWe take rightwards and upwards as positive and origin at top of building
a)At top of trajectory, v_{y}=0
0=30sin(33)-g(t)
t=1.667
s=30sin(33)(1.667)-0.5g(1.667)^{2}
s=13.6m
b)v^{2}=30sin(33)^{2}-2g(-15)
v_{y}=-23.68m/s
v_{y}^{2}+(30cos(33))^{2}=34.55^{2}
v=34.6m/s
c)v_{y}=30sin(33)-g(t)
-23.68=30sin(33)-g(t)
t=4.084s
s_{x}=30cos(33)*4.048=103m
