1. Because the ball was hit horizontally (x-axis-direction) that mean that there is an initial velocity in the x-direction.
2. The question asks what is the smallest velocity needed for the ball to pass through the net.
If you think about it, there will be no vertical initial velocity, (in the y-axis-direction) because the ball was not hit at an angle, it was hit just horizontally, so that means that the initial velocity in the x-direction is what we want to find.
3. Just like vectors, we analyze each direction of motion seperately because in projectile motion (which is what this is) what happens in one direction is independent of the other.
It's very important to understand that we can solve one equation for something like time or velocity, something we do not know in either equation, and then swap this unknown thing for other things like position that we do know.
x \ = \ x_0 \ + \ (v_0_x)t \ + \ \frac{1}{2} a t^2
&
y \ = \ y_0 \ + \ (v_0_y)t \ + \ \frac{1}{2} a t^2
are the two separate equations we'll be using.
4. We use the x-axis direction and see how far we get.
Remember, there is nothing to cause an acceleration in this direction.
Also, we choose the initial position x_0 to be zero.
x \ = \ x_0 \ + \ (v_0_x)t \ + \ \frac{1}{2} a t^2
x \ = \ 0 \ + \ (v_0_x)t \ + \ 0
x \ = \ (v_0_x)t
We know the position x away from the net is 15m. v_0 the initial velocity is what we want.
The only thing we don't know is the time it will take so we'll play with the other equation
and solve for time.
If we do this we can come back to this equation & plug it in.
5. We'll play with the y-axis equation.
There is an acceleration due to gravity of 9.8 m/s^2 downwards.
We'll have to change the a = acceleration sign to a minus to account for this
There is no initial velocity in the y-axis.
y \ = \ y_0 \ + \ (v_0_y)t \ - \ \frac{1}{2} a t^2
y \ = \ y_0 \ + \ 0 \ - \ \frac{1}{2} a t^2
In this equation we can set y_0 [the initial height] to be 2.4m which is the height of the player who hit the ball.
Think about it, if the guy hits the ball horizontally it will move downwards.
Gravity is pulling it downwards.
The question tells us that the net has a height of 0.9m so we want the final height to be equal to this or higher. we'll do what the question asked us to do and set y = 0.9m because that is the minimum.
Solve for time first, then plug in numbers;
y \ = \ y_0 \ + \ 0 \ - \ \frac{1}{2} a t^2
0.9m \ = \ 2.4m \ + \ 0 \ - \ \frac{1}{2} (9.8 m/s^2) t^2
0.9m \ - \ 2.4m \ = \ - \ \frac{1}{2} (9.8 m/s^2) t^2
- \ 1.5 m\ = \ - \ \frac{1}{2} (9.8 m/s^2) t^2
multiply both sides by minus 2 to cancel the minuses & the fraction;
3m \ = \ (9.8 m/s^2) t^2
t^2 \ = \ \frac{3m}{(9.8 m/s^2)}
t^2 \ = \ 0.30612s^2
t = \pm \sqrt{0.30612s^2}
t = \pm 0.5532
So, we have the time (the positive value because it physically can't be a negative time in this situation).
We'll now solve for v_0
x \ = \ (v_0_x)t
(v_0_x) \ = \ \frac{x}{t}
(v_0_x) \ = \ \frac{15m}{0.5532s}
(v_0_x) \ = \ 27.11 m/s
(v_0_x) \ \approx \ 27 m/s
It is perfectly find to round off the 27.11 m/s to 27 m/s.
This may look complicated but it only takes 20 seconds to do once you know it.
I advise you to watch the video to get a better understanding.
Also, I advise you to read this carefully & look at everything, even the units & me using symbols not numbers etc...
Have a goo day
) & derive all of the ones you listed by basic algebra.
