Answer: The initial velocity in the x-axis is 5.08m/s, and in the y-axis is 0m/s
Explanation: Here we know that a projectile is launched at an initial height of 8m, and it travels 6.5 meters before hitting the ground.
First, let's find the time it takes the projectile to reach the ground.
The only force acting on the projectile is gravity force (notice that there is no initial velocity in the y-axis), so the movement equation of the projectile in the y-axis can be written as:
y(t) = -(g/2)t^2 + 8m
where g = 9.8m/s^2 and 8m is the initial height.
Now, we can solve it for t and see how long it takes to the projectile to rech the ground (this happens when y(t) = 0).
0 = -4.9*t^2 + 8m
The solutions are:
[tex]t = \frac{\sqrt{-4*-4.9*8} }{-9.8} = -1.28 s[/tex]
[tex]t = \frac{-\sqrt{-4*-4.9*8} }{-9.8} = 1.28 s[/tex]
Where both solutions are rounded to the nearest hundredth
Where we take the positive solution, because is the one that makes sense.
Now that we know the time that the projectile took to reach the ground, and the distance that it travels in this time, we can see that the initial velocity in the x-axis is:
v = 6.5m/(1.28s) = 5.08m/s
where te velocity is rounded to the nearest hundredth.
