NASA launches a rocket at t=0 seconds. Its height, in meters above sea level, as a function of time is given by h(t) = - 4.91? + 82t + 241.

(a) The rock splashes down when h(t) = 0. Set h(t) = 0 and solve for t using quadratic formula.0 = − 4.9t^2 + 334t + 255t = -0.75, 68.92(Throw away the negative answer, since time can't be negative.)t = 68.92 sec(b) The rock is at it's peak when the slope of it's trajectory is zero. This trajectory is also the tangent line, so finding when the slope of the tangent line is zero is the goal. To do this, take the derivative of h(t) and set h'(t) = 0.h'(t) = -9.8t + 3340 = -9.8t + 334t = 34.08 secUse this time to find the height of the ball h(t), so h(34.08 sec).h(t) = − 4.9(34.08)^2 + 334(34.08) + 255h(t) = 5946.6m And follow me