A) [tex]d=v_0t+\frac{1}{2}at^2[/tex]
0.532 m = 0.202v_0 + ½(-9.81 m/s²)(0.202 s)²
v_0 = 3.62 m/s
B) [tex]d = \frac{1}{2}(v+v_0)t[/tex]
0.532 m = ½(v + 3.62 m/s)(0.202) s
v = 1.65 m/s
C) [tex]v^2=v_0^2+2ad[/tex]
At the highest point v = 0, so
0 = (1.65 m/s)^2 + 2(-9.81 m/s²)d
d = 0.139 m
