Question

Projectile Is shot upward from the surface of Earth with an initial velocity of 136 meters per second. Use the position
function below for free-falling objects. What is its velocity after 3 seconds? After 15 seconds? (Round your answers to one
decimal place.)

s(t) = -4.9t^2 + V0t + S0.

3s =
15s =​

Answer 1

Answer:

$v(3\,s) = 106.6\,\frac{ft}{s}$, $v(15\,s) = -11\,\frac{ft}{s}$

Step-by-step explanation:

The velocity function of the projectile is obtained by deriving the position function:

$v(t) = -9.8\cdot t + 136$

Now, the velocities after 3 and 15 seconds are, respectively:

$v(3\,s) = -9.8\cdot (3) + 136$

$v(3\,s) = 106.6\,\frac{ft}{s}$

$v(15\,s) = -9.8\cdot (15) + 136$

$v(15\,s) = -11\,\frac{ft}{s}$