An object thrown vertically upward from the surface of a celestial body at a vel

Question

An object thrown vertically upward from the surface of a celestial body at a velocity of 30 m/s reaches a height of s-0.6t +30t meters in t seconds. a. Determine the velocity v of the object after t seconds. b. When does the object reach its highest point? c. What is the height of the object at the highest point? d. When does the object strike the ground? e. With what velocity does the object strike the ground? 2 a. The object's velocity after t seconds ism/s (Simplify your answer.) b. It takes the objectseconds to reach its highest point. (Simplify your answer.) c. The object reaches a maximum height ofmeters (Simplify your answer.) d. The object strikes the ground atseconds. (Simplify your answer.) m/s. e. The object strikes the ground with the velocity (Simplify your answer.)

Explanation / Answer

s = -0.6t^2 + 30t

a)
v = ds/dt

v = -0.6(2t) + 30

v = -1.2t + 30 -----> ANS

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b)
s = -0.6t^2 + 30t

Now, this is parabolic, so find vertex

t = -b/(2a)
= -30 / (2*-0.6)
= -30 / -1.2
25

So, 25 seconds to reach the highest point

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c)
What is the max ht?

s = -0.6t^2 + 30t

s = -0.6(25)^2 + 30(25)

375 meters ----> ANS

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d)
It strikes grnd when s = 0

-0.6t^2 + 30t = 0

-0.6t(t - 50) = 0

t = 0 or t = 50

So, it struck ground
after 50 second ----> AMS

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e)
v = -1.2t + 30

But it strikes grnd at t = 50, so plug that in :
v = -1.2 * 50 + 30

v = -30 meter per second -----> ANS

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