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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 =​

Respuesta :

xero099

Answer:

[tex]v(3\,s) = 106.6\,\frac{ft}{s}[/tex], [tex]v(15\,s) = -11\,\frac{ft}{s}[/tex]

Step-by-step explanation:

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

[tex]v(t) = -9.8\cdot t + 136[/tex]

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

[tex]v(3\,s) = -9.8\cdot (3) + 136[/tex]

[tex]v(3\,s) = 106.6\,\frac{ft}{s}[/tex]

[tex]v(15\,s) = -9.8\cdot (15) + 136[/tex]

[tex]v(15\,s) = -11\,\frac{ft}{s}[/tex]

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