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

Step-by-step explanation:

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

`y=y_(o)+V_(o)sin \theta t-(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-(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.