Question

While standing on a 4-foot ladder, a grapefruit is tossed straight up with an initial velocity of 40 fu/sec. The initial position ofthe grapefruit is 5 feet above the ground when it is released. Its height at time t is given by y = h(t) = -16t^2 + 40t + 5.a) How high does it go before returning to the ground? Round time to 2 decimal places to compute height.____feet.b) How long does it take the grapefruit to hit the ground? Round time to 3 decimal places.____ seconds

Answer 1

a. 30 feet

b. 2.368 s

Step-by-step explanation:

a)

The initial position of the grapefruit is 5 feet above the ground when it is released. Its height at time t is given by

y = h(t) = -16t^2 + 40t + 5.

If we plot this function on a graph, we would get a parabola. The vertex of the parabola represents the maximum height reached by the grapefruit before reaching the ground. We would calculate the x coordinate of the vertex of the parabola by applying the formula,

x = - b/2a

Recall, the standard form of a quadratic equation,

y = ax^2 + bx + c

By comparing both equations,

a = - 16

b = 40

c = 5

Thus,

x = - 40/2* - 16 = - 40/- 32 = 1.25

We would find the y coordinate of the vertex by substituting t = 1.25 into the original equation. We have

h(1.25) = -16(1.25)^2 + 40(1.25) + 5

h(1.25) = - 25 + 50 + 5

h(1.25) = 30

The height is 30 feet

b) When the grapefruit reaches the ground, h = 0

Substituting h = 0 into the original equation, we have

-16t^2 + 40t + 5 = 0

We would apply the quadratic formula which is expressed as

`\begin{gathered} x\text{ = }\frac{-\text{ b }\pm√(b^2-4ac)}{2a} \\ x\text{ = }\frac{-\text{ 40 }\pm\sqrt{40^2-4(-\text{ 16}*5)}}{2*-16} \\ x\text{ = }\frac{-\text{ 40 }\pm√(1280)}{-\text{ 32}} \\ x\text{ = }\frac{-\text{ 40 + 35.777}}{-\text{ 32}}\text{ or x = }\frac{-\text{ 40 - 35.777}}{-\text{ 32}} \end{gathered}`

t = 0.132 or t = 2.368

Since it takes 1.25s to reach maximum height, the reasonable time for it to reach the ground is 2.368 s