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Sagot :
(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.
Projectile velocity at the highest point of trajectory
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
Maximum height of the projectile
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
Learn more about projectile here: https://brainly.com/question/24216590
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