1. A baseball is thrown from the top of a building. The height of the ball is given by the equation h=-5t^2+5t+30 where h is the height of the ball in metres after t seconds?

Question

(a) Sketch a graph of this relationship by first finding the x–intercepts (zeros) and then using the

 x-intercepts to find the vertex.

(b) When will the ball hit the ground?

(c) What is the maximum height of the ball and when does it occur?

(d) What is the starting height of the ball?

(e) What is the height of the ball after 2 seconds?

Answer 1

h(t) = -5t^2 +5t + 30 is a downward opening parabola with y intercept = 30

-5t^2 + 5t + 30 = 0

t^2 -t - 6 = 0

(t-3)(t+2) = 0

t = -2, 3 are the two x intercepts. in 3 seconds the ball hits the ground

it reaches maximum height at the midpoint between t = -2 and t =3, which is (-2+3)/2 = 1/2 second

(at 1/2 second h(1/2) = -5(1/2)^2 + 5(1/2) + 30 = -5/4 + 5/2 + 30 = 31.25 meters

vertex is (0.5, 31.25)

alternative method is take the derivative and set equal to zero

h'(t) = -10t +5 = 0

t =5/10 = 1/2 second at maximum height

starting height is 30 meters, the constant term in the height formula

starting velocity is 5 meters per second, the coefficient of the linear term

acceleration = minus ten meters per second per second, twice the coefficient of the quadratic term

in 2 seconds the height = h(2) = -5(2)^2 + 5(2) + 30 = -20 +10 + 30 = 20 meters high

(the problem rounded off the actual effect of gravity, which is closer to -4.9t^2, not -5t^2, as deceleration at sea level = -9.8 meters per second per second, and higher up, it's slightly less, not more. Had the problem involved feet, it would have had -16t^2. -5t^2 indicates deceleration is measured in meters per second per second)