Question

a projectile is shot upward from the surface of earth with an initial velocity of 89 meters per second. use the position function below for free-falling objects. what is its velocity after 2 seconds? after 10 seconds? (round your answers to one decimal place.)

Answer 1

the velocity of the projectile after 2 seconds is 70.4 m/s and after 10 seconds is -92.1 m/s

Step-by-step explanation:

The position function for a free-falling object is:

x(t) = x0 + v0*t - (1/2)gt^2

where x(t) is the position of the object at time t, x0 is the initial position of the object, v0 is the initial velocity of the object, and g is the acceleration due to gravity (9.81 m/s^2).

To find the velocity of the object after 2 seconds, we can use the derivative of the position function:

v(t) = dx/dt = v0 - g*t

Plugging in t=2 and the given initial velocity of 89 m/s, we get:

v(2) = 89 m/s - (9.81 m/s^2) * 2 s = 70.38 m/s

To find the velocity of the object after 10 seconds, we can again use the derivative of the position function:

v(t) = v0 - g*t

Plugging in t=10 and the given initial velocity of 89 m/s, we get:

v(10) = 89 m/s - (9.81 m/s^2) * 10 s = -92.1 m/s

So, the velocity of the projectile after 2 seconds is 70.4 m/s and after 10 seconds is -92.1 m/s.