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The height (in meters) of a projectile shot vertically upward from a point 3 m above ground level with an initial velocity of 23.5 m/s is
h = 3 + 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.

Respuesta :

Answer:

a)

at t = 2s

[tex]v_y = 3.9 m/s[/tex]

b)

at t = 4 s

[tex]v_y = -15.7 m/s[/tex]

Explanation:

As we know that the velocity of the object is given as rate of change in its position

so we will have

[tex]v_y = \frac{dh}{dt}[/tex]

now we will have

[tex]v_y = \frac{d}{dt}(3 + 23.5 t - 4.9 t^2)[/tex]

so we will have

[tex]v_y = 23.5 - 9.8 t[/tex]

a)

at t = 2s

[tex]v_y = 3.9 m/s[/tex]

b)

at t = 4 s

[tex]v_y = -15.7 m/s[/tex]

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