given horizontal distance=68 m
angle with respect to horizontal, 35.8
and time object spends in the air is 3.0 seconds.
thanks.
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: .$\displaystyle v_o \;=\;\frac{4.9(3^2)}{3\sin35.8^o} \;\approx\;\boxed{25.13\text{ m/sec}}$