Solution.
x=86t; y=96t−4.9t2;
Since the trajectory of motion is a parabola, the motion of the projectile is described by the formulas:
x=υ0xt; y=υ0y t+ayt2/2;
a) ax=0;ay=−9.8m/s2;
b) υx=86m/s;υy=96−9.8t;
c) υ0=υx2+υy2; υ0=862+962=128.9m/s;
d) υx=86m/s;υy=96−9.8⋅5=47m/s;
e) hm=−υ0y2 /2⋅ay;
hm=−962/2⋅(−9.8)=470.2m;
f) L=υ0xtm; tm=2⋅(−υ0y/ay); tm=2⋅(−96/−9.8)=19.6s;
L=86⋅19.6=1684.9m;
Answer: a) ax=0;ay=−9.8m/s2;
b) υx=86m/s; υy=96−9.8t;
c) υ0=128.9m/s;
d) υx=86m/s;υy=47m/s;
e)hm=470.2m;
f)L=1684.9m;
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