Answer:
The horizontal distance is 2.41 mts
Explanation:
For this problem, we will use the formulas of parabolic motion.
[tex]Y=Yo+Voy*t+\frac{1}{2}*a*t^2\\Vx=V*cos(\alpha)\\Vy=V*sin(\alpha)[/tex]
We need to find the time of the whole movement (t), for that we will use the first formula:
we need the initial velocity for that:
[tex]Vy=4.4*sin(24^o)\\Vy=1.79m/s[/tex]
so:
[tex]0=0.70+1.79*t+\frac{1}{2}*(-9.8)*t^2[/tex]
now we have a quadratic function, solving this we obtain two values of time:
t1=0.60sec
t2=-0.234sec
the obvious value is 0.60sec, we cannot use a negative time.
Now we are focusing on finding the horizontal distance.
the movement on X is a constant velocity motion, so:
[tex]x=Vx*t[/tex]
[tex]Vx=4.4*cos(24^o)=4.02m/s\\[/tex]
so:
[tex]x=4.02*(0.60)=2.41m[/tex]
