Question

A cannon ball is shot from a cannon with an upward velocity of 40 feet per second. The equation that gives height h, in feet, of the ball at time t, in seconds h(t) = -16t^2 + 40t + 1.5.

a. What is the height of the ball after 2 seconds?
b. What is the maximum height reached by the ball?
c. When does the ball meet its maximum height?
d. How long will it take before the ball hits the ground?

Answer 1

a. The height of the ball after 2 seconds is 17.5 feet. b. The maximum height reached by the ball is 26.5 feet. c. The ball meets its maximum height at 1.25 seconds. d. The ball hits the ground after approximately 3.79 seconds.

Step-by-step explanation:

a. To find the height of the ball after 2 seconds, we substitute t = 2 into the equation h(t) = -16t^2 + 40t + 1.5:

h(2) = -16(2)^2 + 40(2) + 1.5
= -64 + 80 + 1.5
= 17.5 feet

b. To find the maximum height reached by the ball, we need to find the vertex of the parabolic equation. The vertex is given by the t-coordinate of the vertex, which is -b/2a, where a, b, and c are the coefficients of the equation h(t) = -16t^2 + 40t + 1.5:

t = -40/(-2*16)
= 1.25 seconds

Substituting t = 1.25 into the equation, we can find the maximum height:
h(1.25) = -16(1.25)^2 + 40(1.25) + 1.5
= -25 + 50 + 1.5
= 26.5 feet

c. The ball meets its maximum height at t = 1.25 seconds.

d. To find how long it takes for the ball to hit the ground, we can set h(t) = 0 and solve for t:
0 = -16t^2 + 40t + 1.5
Using the quadratic formula, we get t ≈ 3.79 seconds and t ≈ 0.54 seconds. The ball hits the ground after approximately 3.79 seconds.