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Svetach [21]
2 years ago
6

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
Physics
1 answer:
Burka [1]2 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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2 years ago
If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
Butoxors [25]
(a) 3.56 m/s 
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For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
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