physics question; how to find the velocity

• Oct 25th 2007, 11:57 AM
beetz
physics 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, 12:20 PM
Soroban
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: . $h \;=\;h_o + (v_o\sin\theta)t - 4.9t^2$

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

• Oct 25th 2007, 12:29 PM
beetz
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, 03:02 PM
CaptainBlack
Quote:

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