given horizontal distance=68 m

angle with respect to horizontal, 35.8

and time object spends in the air is 3.0 seconds.

thanks.

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- Oct 25th 2007, 10:57 AMbeetzphysics question; how to find the velocity
given horizontal distance=68 m

angle with respect to horizontal, 35.8

and time object spends in the air is 3.0 seconds.

thanks. - Oct 25th 2007, 11:20 AMSoroban
Hello, beetz!

Quote:

Given: horizontal distance = 68 m

angle with respect to horizontal: 35.8°

time object spends in the air: 3.0 seconds.

Find the initial velocity.

We're expected to know this formula: .$\displaystyle h \;=\;h_o + (v_o\sin\theta)t - 4.9t^2$

. . where: .$\displaystyle \begin{array}{ccc}h_o & = & \text{initial height} \\

v_o & = & \text{initial velocity} \\

\theta & = & \text{angle of elevation} \\

h & = & \text{height of object} \\

& & \text{at time }t

\end{array}$

We are given: .$\displaystyle h_o = 0,\;\theta = 35.8^o$

. . Hence, we have: .$\displaystyle h \;=\;(v_o\sin35.8^o)t - 4.9t^2$

The object is on the ground $\displaystyle (h = 0)$ when $\displaystyle t = 0$ and $\displaystyle t = 3.$

So we have: .$\displaystyle (v_o\sin35.8^o)3 - 4.9(3^2) \;=\;0$

Therefore: .$\displaystyle v_o \;=\;\frac{4.9(3^2)}{3\sin35.8^o} \;\approx\;\boxed{25.13\text{ m/sec}}$

- Oct 25th 2007, 11:29 AMbeetz
thank you so much.

the equation in my notes is a bunch of mumbo jumbo.

though, almost everything in physics seems like a bunch of mumbo jumbo to me. :p - Oct 25th 2007, 02:02 PMCaptainBlack