Question

If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table at a height of 1.1 m, how far horizontally will it hit the ground?

Answer 1

Answer: 1.12 m

Explanation:

This situation is related to parabolic motion, hence we can use the following equations:

$y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2}$ (1)

$x=V_{o} cos \theta t$ (2)

Where:

$y=0 m$ is the ball final height (when it hits the ground)

$y_{o}=1.1 m$ is the ball initial height

$V_{o}=2.2 m/s$ is the initial velocity

$\theta=30\°$ is the angle at which the ball was launched

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the horizontal distance the ball travels

Rewriting (1) with the given values:

$0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2}$ (3)

Multiplying all the eqquation by -1 and rearranging:

$4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0$ (4)

So, since we have a quadratic equation here (in the form of$0=at^{2}+bt+c$,  we will use the quadratic formula to find  $t$:  

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$   (5)

Where $a=4.9$, $b=-1.1$, $c=-1.1$  

Substituting the known values and choosing the positive result of the equation, we have:  

$t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}$  

$t=0.59 s$ (6)

Now, substituting (6) in (2):

$x=(2.2 m/s)(cos 30\°)(0.59 s)$ (7)

$x=1.12 m$ (8) This is the horizontal distance at which the ball hits the ground.