(a) The projectile's velocity at the highest point of its trajectory is 51.96 m/s.
(b) The maximum height reached by the projectile is 45.92 m.
At the highest point of trajectory the vertical component of the velocity will be zero will the horizontal component will remain constant.
Vxi = Vxf = Vcosθ = 60 x cos30 = 51.96 m/s
The maximum height reached by the projectile is calculated as follows;
H = u²sin²θ/2g
H = (60² x (sin30)²)/(2 x 9.8)
H = 45.92 m
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