the solutions to Midterm 1 - Loyola University Chicago

Problem 3.(12 points total) The height of a projectile fired straight up in the air, as a function oftime since the firing, is given byh(t) = −5t 2 + 40t.a.(5 points)Find the time when the projectile is at its highest position.Solution: The projectile is at the highest position when its velocity is 0. Nowand v(t) = 0 when t = 4.v(t) = h ′ (t) = −10t + 40Another way to solve this is to note that h(t) = −5t 2 + 40t = −5(t 2 − 8t) = −5t(t − 8) = 0 whent = 0 or t = 8. The highest point on the parabola is when t is half way between 0 and 8, which is att = 4.b.(7 points)Find the velocity of the particle at the moment it hits the ground.Solution: The particle hits the ground when h(t) = 0. Solve h(t) = −5t 2 + 40t = −5(t 2 − 8t) =−5t(t − 8) = 0 to get t = 0 or t = 8. The particle is fired at t = 0 so it hits the ground when t = 8.Velocity is the derivative of height, sov(t) = h ′ (t) = −10t + 40andv(8) = −40.