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28 votes
28 votes
Starting at an initial height of 364 feet, an object is projected

upward at 60 feet per second. In how many seconds will it
reach the ground? I’m

User Gdavis
by
2.5k points

1 Answer

7 votes
7 votes

Step-by-step explanation:

4.5 s, 324 ft

Step-by-step explanation:

The object is projected upward with an initial velocity of

v_0 = 144 ft/sv

0

=144ft/s

The equation that describes its height at time t is

s(t) = -16t^2 + 144 ts(t)=−16t

2

+144t (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

v(t) = s'(t) = -32t +144v(t)=s

(t)=−32t+144 (2)

At the maximum heigth, the vertical velocity is zero:

v(t) = 0

Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

\begin{gathered}0=-32t+144\\t=\frac{144}{32}=4.5 s\end{gathered}

0=−32t+144

t=

32

144

=4.5s

And by substituting this value into eq.(1), we also find the maximum height:

s(t) = -16(4.5)^2+144(4.5)=324 fts(t)=−16(4.5)

2

+144(4.5)=324ft

used this way i even don't know sorry:))

User Algiz
by
2.7k points