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h=-16t^(2) + vt + k The equation above...

`h=-16t^(2) + vt + k`
The equation above gives the height h, in feet, of a ball t seconds after it is thrown straight up with an initial speed of v feet per second from a height of k feet. Which of the following gives v in terms of h, t, and k ?

A

`v=h + k - 16t`

B

`v=(h-k + 16)/t`

C

`v=(h+k)/t - 16t`

D

`v=(h-k)/t + 16t`

Text Solution

AI Generated Solution

The correct Answer is:
To solve for \( v \) in terms of \( h \), \( t \), and \( k \) from the equation \( h = -16t^2 + vt + k \), we can follow these steps: ### Step 1: Write down the original equation We start with the given equation: \[ h = -16t^2 + vt + k \] ### Step 2: Rearrange the equation to isolate \( vt \) To isolate \( vt \), we can move \( -16t^2 \) and \( k \) to the left side: \[ vt = h + 16t^2 - k \] ### Step 3: Solve for \( v \) Now, we can solve for \( v \) by dividing both sides by \( t \): \[ v = \frac{h + 16t^2 - k}{t} \] ### Step 4: Simplify the expression We can simplify the expression further: \[ v = \frac{h - k}{t} + 16t \] ### Final Expression Thus, the final expression for \( v \) in terms of \( h \), \( t \), and \( k \) is: \[ v = \frac{h - k}{t} + 16t \]
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