
Kenneth S. answered 07/29/17
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Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
a) the max heght is at the vertex, so t = -b/(2a) = 30/6 = 5 seconds. so max ht is s(5); you can compute that.
b) velocity when hits ground will be s'(10). differentiate s(t) & substitute t=10 for this computation. why t=10? because of symmetry of the parabola...5 sec to vertex, 5 sec back down to ground since 'launch' was from ground level.
c) a(t) = s"(t) = -6 m/sec2