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Question:

The height of a body is described by:

$h = \frac{\ln\left(t^2-B\right)}{a}$

where is the time in seconds and and are constants.

a) Write in terms of , and ℎ.
b) Determine $\frac{ℎ}{}$
c) Given that $\alpha = 0.2$ and $ = 9$, calculate the positive value of such that $\frac{ℎ}{}=50$

My attempt:

Part a. I have $t = \sqrt{\mathrm{e}^{ha}+B}$

Part b. using the chain rule I've differentiated to get :

$\frac{ℎ}{}$ $=\dfrac{2t}{a\left(t^2-B\right)}$

Now for part c. I need to use:

$50$ $=\dfrac{2t}{0.2\left(t^2-9\right)}$

Can someone show me how to rearrange to find t here, or If i have went wrong somewhere please?

Im looking for an answer of 3.102 seconds

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1 Answers1

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You have a linear and a quadratic power on $t$. The obvious way to start off, thus, is trying to form a quadratic equation in $t$.

On rearranging we get: $5(t^2-9)=t$ which is equal to $5t^2-t-45=0$ - a quadratic equation!

Can you solve it now?

Gaurang Tandon
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