Answer to Question #227566 in Physics for arte08

Question

A stone is thrown vertically upward from the top of the building. If the equation of the motion of the stone is 𝑠(𝑡) = −5𝑡² + 30𝑡 + 200 , where 𝑠 is the directed distance from the ground in meters and 𝑡 is in seconds.

a. What is the height of the building?

b. What is the average velocity on the time interval [1,3]

c. What is the instantaneous velocity at time t=1?

d. at what time the stone will hit the ground?

e. What is the instantaneous velocity of the stone upon impact?

Answer 1

a. At $t = 0$ the value of $s(t)$ gives the height of the building. Thus:

$$h = s(0) = 200m$$

b. The distance from the ground at $t = 1$ is:

$$s(1) = -5\cdot 1^2 + 30\cdot 1 + 200 = 225m$$

The distance from the ground at $t = 3$ is:

$$s(3) = -5\cdot 3^2 + 30\cdot 3 + 200 = 245m$$

The average velocity is:

$$v_{av} = \dfrac{s(3) - s(1)}{3s - 1s} \approx 6.7m/s$$

c. The instantaneous velocity is:

$$v(t) = s'(t) = -10t+30\\ v(1) = -10+30 = 20m/s$$

d. The stone will hit the ground when $s(t) = 0.$ Solving the following quadratic equation for $t$, obtain:

$$-5t^2 + 30t + 200 = 0\\ t = 10s$$

c. The instantaneous velocity at this time is:

$$v(10) = -10\cdot 10 + 30 = -70m/s$$