Answer to Question #219236 in Algebra for city girl fran

Question

In the equations, tv, and hare defined as follows:

.

t=the number of seconds after a sensor is dropped

v=the speed, in feet per second, of a sensor seconds after it is dropped

h-the height above the ground, in feet, of sensor (seconds after it is dropped

THE FINA

Use the scientist's equations to answer the questions below.

he exactly

a. What is the speed, in foot per second, of a sensor S seconds after it is dropped? Show or explain

how you got your answer

ETE ANYM

b. How many seconds after a sensor is dropped will its speed be 384 feet per second? Show or

explain how you got your answer,

pverv at

What is the height above the ground. In feet of a sensor 5 seconds after it is dropped? Show or

explain how you got your answer.

Answer 1

Kinematic equations for object in free-fall where acceleration $a=-g$

$$h(t)=h_0+v_0t-\dfrac{gt^2}{2}$$

$$v(t)=v _0-gt$$

$g=32 \ ft/s^2, v_0=0, h_0=400\ ft$

Let

$$h(t)=400-\dfrac{32t^2}{2}$$

$$v(t)=-32t$$

a.

$$v(S)=-32S\ ft/s$$

b.

If We take $v(t_1)=-32t_1=-384,$ then $t_1=12\ s$ and

$$h(12)=400-\dfrac{32(12)^2}{2}=-1904 (ft)$$

The result is impossible, since $h(t)\geq0$ during the flight.

So we take

$$v(t_1)=-32t_1=-38.4 ft/s$$

$$t_1=1.2\ s$$

c.

$$h(5)=400-\dfrac{32(5)^2}{2}$$

$$h(5)=0\ ft$$

The sensor will reach the ground 5 seconds after it is dropped.