Answer: 1.56 m
Explanation:
This situation is a good example of the projectile motion or parabolic motion, and the main equations that will be helpful in this situations are:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Where:
[tex]x[/tex] is the horizontal distance
[tex]V_{o}=3.5 m/s[/tex] is the initial speed
[tex]\theta=50\°[/tex] is the angle at which the venom was shot
[tex]t[/tex] is the time since the venom is shot until it hits the ground
y-component:
[tex]y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}[/tex] (2)
Where:
[tex]y_{o}=0.5m[/tex] is the initial height of the venom
[tex]y=0[/tex] is the final height of the venom (when it finally hits the ground)
[tex]g=-9.8m/s^{2}[/tex] is the acceleration due gravity
Knowing this, let's begin:
First we have to find [tex]t[/tex] from (2):
[tex]0=0.5 m+3.5m/s sin(50\°) t+\frac{-9.8m/s^{2}t^{2}}{2}[/tex] (3)
Rearranging (3):
[tex]-4.9 m/s^{2} t^{2} + 2.681 m/s t + 0.5 m=0[/tex] (4)
This is a quadratic equation (also called equation of the second degree) of the form [tex]at^{2}+bt+c=0[/tex], which can be solved with the following formula:
[tex]t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex] (5)
Where:
[tex]a=-4.9[/tex]
[tex]b=2.681[/tex]
[tex]c=0.5[/tex]
Substituting the known values:
[tex]t=\frac{-2.681 \pm \sqrt{(2.681)^{2}-4(-4.9)(0.5)}}{2(-4.9)}[/tex] (6)
Solving (6) we find the positive result is:
[tex]t=0.694 s[/tex] (7)
Substituting (7) in (1):
[tex]x=3.5 m/s cos(50\°) (0.694 s)[/tex] (8)
Finally:
[tex]x=1.56 m[/tex] (9)
