# Thread: physics question; how to find the velocity

1. ## 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.

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

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

3. 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.

4. 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