ALTERNATIVE 1
uy=64sin7/7.80<ms−1≫
decrease in height =7.80×0.187+21×9.81×0.1872/1.63 «m»
final height = « 2.80-1.63 » =1.1 / 1.2 «m»
«higher than net so goes over»
ALTERNATIVE 2
vertical distance to fall to net «=2.80-0.91 »=1.89 «m»
time to fall this distance found using <1.89=7.8t+21×9.81×t2 »
t=0.21 «s»
0.21 «s» > 0.187 «s»
«reaches the net before it has fallen far enough so goes over»
Other alternatives are possible