physics question; how to find the velocity

**beetz** · October 25th 2007, 10:57 AM

given horizontal distance=68 m
angle with respect to horizontal, 35.8
and time object spends in the air is 3.0 seconds.

thanks.

**Soroban** · October 25th 2007, 11:20 AM

Hello, beetz!

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: . $h \;=\;h_o + (v_o\sin\theta)t - 4.9t^2$

. . where: . $\begin{array}{ccc}h_o & = & \text{initial height} \\ v_o & = & \text{initial velocity} \\ \theta & = & \text{angle of elevation} \\ h & = & \text{height of object} \\ & & \text{at time }t \end{array}$

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

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

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

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

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

**beetz** · October 25th 2007, 11:29 AM

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.

**CaptainBlack** · October 25th 2007, 02:02 PM

Originally Posted by beetz

given horizontal distance=68 m
angle with respect to horizontal, 35.8
and time object spends in the air is 3.0 seconds.

thanks.

You need to provide more detail otherwise we are just guessing at your question

RonL

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