SOLUTION: Suppose a cannon shoots a cannon ball into the air so that the initial upward velocity of the cannon ball is 320 feet per second.Assume the cannon ball is shot from ground level.
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Question 65312: Suppose a cannon shoots a cannon ball into the air so that the initial upward velocity of the cannon ball is 320 feet per second.Assume the cannon ball is shot from ground level.
H=-1/2*gt^+vt+h where g is the acceleration due to gravity, and on the earth, equals 32ft/sec^ (English units) or 9.8m/sec^(metric units,
v is the initial upward velocity of the object
h is the initial height above the ground
the question:
how high will the cannon ball be in 7 seconds?
and how do you figure it out?
i tried -16x^+320*7 because i thought this is what the solver was telling me to do...my answers were 2256,2224,2128,and 2352. None of these appear to be correct. I answered question #1 by guessing: in 5 seconds, the rocket will be 1200 feet above ground. The solution I came up with was 320*5=1600, 16*5=80,then add 320+80=400, then subtract 400 from 1600 to get 1200feet, which was the correct answer , but for question #2, i am wrong if i use the same pattern.
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
The equation that relates the height (as a funtion of time) of an object propelled upwards at an initial velocity of Vo from an initial height of Ho is:
Note the in the first term. Substituting t = 7, you get:
Simplify.
The cannon ball will reach a height of 1,456 feet in 7 seconds.
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