A ball is shot at a velocity of $10.0\ m/s$ at $40.0^\circ$ above the horizontal. How far away does it land?
I know that the horizontal displacement equals time*horizontal velocity where here the horizontal velocity is $8\ m/s$. I don't know how to find the time.
Use Components!
$sin(40) (10)=6.43$
now that is the vertical component of velocity.
$d=v_i t +\frac{1}{2}a t^2$
when the ball lands, $d=0$
$0=6.435t-9.81 (0.5)(t^2)$ ($a$ is our acceleration and is equal to gravity)
which gives $t=0$ or $1.31$.
Ignore $t=0$ as that is when we are launching are projectile and take the other time.
You should set up some parametric equations.
$x = v_0\cos \theta\ t\\ y = v_0\sin \theta\ t - \frac 12 gt$
From here, you can either find the values of $t$ where $y = 0$ and then find $x(t)$
Or you can make an equation for $y$ in terms of $x$ and find $x$ without solving for $t.$
The movement equations are given by \begin{align*} \begin{cases} \displaystyle y(t) = v_{0}\sin(\theta)t - \frac{gt^{2}}{2}\\\\ \displaystyle x(t) = v_{0}\cos(\theta)t \end{cases} \end{align*}
When the ball reaches the ground, we have $y(t) = 0$, that is to say \begin{align*} y(t) = v_{0}\sin(\theta)t - \frac{gt^{2}}{2} = 0 \Longleftrightarrow t_{1} = 0\quad\vee\quad t_{2} = \frac{2v_{0}\sin(\theta)}{g} \end{align*}
Finally, the displacement on the $x$-axis is given by \begin{align*} x(t_{2}) = \frac{v^{2}_{0}\sin(2\theta)}{g} \end{align*}
Then plug in the given values in order to obtain the numerical answer.
Use the facts that the vertical component of the projectile’s velocity is zero at the top of its arc and that this occurs halfway through its flight to derive a very simple linear equation for $t$.
