SOLUTION: Find the height, of a ball, above ground using the following information.
wait 2 seconds, initial velocity is 55 ft/sec, initial height above ground
is 2 ft
Question 1 answers
Algebra.Com
Question 193830: Find the height, of a ball, above ground using the following information.
wait 2 seconds, initial velocity is 55 ft/sec, initial height above ground
is 2 ft
Question 1 answers
40 ft
42 ft
46 ft
48 ft
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Technically speaking, you cannot answer this question from the information given. That is because you haven't actually given the initial velocity. Yes, I know that you said the initial velocity is 55 ft/sec, but that is only part of the story. 55 ft/sec is a speed, i.e. a scalar quantity. Velocity is a vector quantity and requires that you specify both the magnitude (speed) and the direction. 55 ft/sec North, 55 ft/sec horizontally, and 55 ft/sec West at a 45 degree angle to the horizontal, are all very different velocities.
Having said all that, I suspect that you really meant that the initial velocity is 55 ft/sec up. So, let's solve the problem using that assumption.
The acceleration due to the force of gravity on an object near the earth's surface is
Note the minus sign because the accelleration vector is opposite in direction to our velocity vector.
Integrating with respect to time gives the instantaneous velocity at time t:
 = \int\,g\,dt = \int\,-32 \text{ \frac{ft}{sec^2}\,dt = -32t + C \text{ \frac{ft}{sec}})
Where, for this problem, the constant of integration, C, is the initial velocity, and we will call this 
. (If the velocity was in a direction other than straight up, then C would be the vertical component of the initial velocity)
Then integrating the velocity function with respect to time gives the instantaneous height at time t:
 = \int\,V(t) = \int\,-32t + v_o \text{ \frac{ft}{sec}} = -16t^2 + v_ot + C \text{ ft})
Where the constant of integration is the initial height which we will call 
. Hence the final function is:
 = -16t^2 + v_ot + h_o \text{ ft})
And substituting the given initial values:
 = -16t^2 + 55t + 2 \text{ ft})
Now the problem is to find )
 = -16(2)^2 + 55(2) + 2 \text{ ft})
The only thing left is a little arithmetic.
John
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!
Edwin's solution:
I think you meant "initial velocity 55 ft/sec"!
The formula is
Where
is the initial height 2 ft
is the initial velocity 55 ft/sec
is the time you "freeze" the ball 2 seconds.
ft.
Edwin
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