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3 years ago

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 23

.5 m/s is h = 2 + 23.5t − 4.9t2 after t seconds. (Round your answers to two decimal places.) (a) Find the velocity after 2 s and after 4 s. v(2) = m/s v(4) = m/s

1 answer:

Burka [1]3 years ago

8 0

Answer:

a) $v(2) = 3.9\,\frac{m}{s}$ , b) $v(4) = -15.7\,\frac{m}{s}$

Explanation:

a) The equation for vertical velocity is obtained by deriving the function with respect to time:

$v(t) = 23.5 -9.8\cdot t$

The velocities at given instants are, respectivelly:

$v(2) = 3.9\,\frac{m}{s}$

$v(4) = -15.7\,\frac{m}{s}$

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