A ball is thrown up from a height of 50m above the ground. Its height after t seconds is given by the function h(t)=50+15t-5t^2

Question

A) what is the balls acceleration at t=3.5s?

B) what is the maximum height the ball reaches?

Answer 1

A) Velocity v(t)= h'(t) = 15 - 10t;

acceleration a(t) = v'(t) = - 10 m/s² = const; a(3.5) = - 10 m/s².

B) At maximum height v(t) = 15 - 10t = 0; t = 1.5 s

max h(t) = h(1.5) = 50 + 15·1.5 - 5·1.5² = 61.25 m