SOLUTION: Use the position function
s(t) = -16t^2 + v_0t + s_0
v0 = initial velocity,
s0 = initial position,
t = time
to answer the following qu
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Question 994201: Use the position function
s(t) = -16t^2 + v_0t + s_0
v0 = initial velocity,
s0 = initial position,
t = time
to answer the following question.
You throw a ball straight up from a rooftop 170 feet high with an initial velocity of 32 feet per second. During which time period will the ball's height exceed that of the rooftop?
ANSWER: Between ...?? and ...?? seconds.
Found 3 solutions by stanbon, josmiceli, MathLover1:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Use the position function
s(t) = -16t^2 + v_0t + s_0
v0 = initial velocity,
s0 = initial position,
t = time
to answer the following question.
You throw a ball straight up from a rooftop 170 feet high with an initial velocity of 32 feet per second. During which time period will the ball's height exceed that of the rooftop?
===========
-16t^2 + 32t + 170 > 170
-----
-16t^2 + 32t > 0
-----
-16t(t-2) > 0
----
ANSWER: Between 0 and 2 seconds.
----
Cheers,
Stan H.
-------
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
You want to know the values of when
When ,
So, is the initial time
Now you have to find when theball is
coming down and is back to the 170 ft level
The ball exceeds the height of the rooftop
between 0 and 2 sec
--------------------
Here's a plot of the ball's height and time in sec
Answer by MathLover1(20849) (Show Source): You can put this solution on YOUR website!
,
,
given:
=>the ball's height exceed that of the rooftop
.
solutions:
and
so between and at the ball’s height exceed
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