Projectile distance

Question

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.

Answer 1

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.

Answer 2

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 $t.$

Answer 3

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.

Answer 4

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 $t$.