Chapter 6.4, Problem 3E

Reminder Round all answers to two decimal places unless otherwise indicated.

Looking Up The constant $g-32\text{feet}\text{per}\text{second}\text{per}\text{second}$ is the downward acceleration due to gravity near the surface of the Earth. If we stand on the surface of the Earth and locate objects using their distance up from the ground, then the positive direction is up, so down is the negative direction. With this perspective, the equation of change in velocity for a freely falling object would be expressed as

$\frac{dV}{dt}=-g$.

(We measure upward velocity $V$ in feet per second and time $t$ in seconds.) Consider a rock tossed upward from the surface of the Earth with an initial velocity of $40\text{feet}\text{per}\text{second}$ upward.

a. Use a formula to express the velocity $V-V\left(t\right)$ as a linear function. (Hint: You get the slope of $V$ from the equation of change. The vertical intercept is the initial value.)

b. How many seconds after the toss does the rock reach the peak of its flight? (Hint: What is the velocity of the rock when it reaches its peak?)

c. How many seconds after the toss does the rock strike the ground? (Hint: How aces me time it takes for the rock to rise to its peak compare with the time it takes for it to fall hack to the ground?)