Answer:
a
[tex]v = 5.39 \ m/s[/tex]
b
Horizontal component
[tex]v_x = 5.00 \ m/s[/tex]
vertical component
[tex]v_y = - 2.0 \ m/s[/tex]
c
[tex]t = 0.921 \ s[/tex]
d
[tex]d = 4.605 \ m [/tex]
Explanation:
Generally from the question we can deduce that he initial velocity of the cork, as seen by an observer on the ground in terms of the x unit vector is
[tex]v_x = 5.00 \ m/s[/tex] due to the fact that the cork is moving horizontally
Generally from the question we can deduce that the vertical and horizontal components of the initial velocity is
[tex]v_y = - 2.0 \ m/s[/tex] due to the fact that the balloon is moving downward which is the negative which will also cause the cork to move vertically with the balloon speed
Generally the initial velocity (magnitude and direction) of the cork, as seen by an observer on the ground is mathematically represented as
[tex]v = \sqrt{ v^2 _x + v^2 _y }[/tex]
[tex]v = \sqrt{ 5^2 + (-2)^2 _y }[/tex]
[tex]v = 5.39 \ m/s[/tex]
Generally the initial direction of motion as seen by the same observer is mathematically represented as
[tex]\theta = tan^{-1}[\frac{2}{5} ][/tex]
[tex]\theta = 21.80^o[/tex]
Generally the time taken by the cork in the air before landing is mathematically represented as
[tex]D = ut + \frac{1}{2} g t^2[/tex]
So D = 6 \ m from the question
g = 9.8 \ m/s^2
u = [tex]v_y[/tex] = 2 m/s this because we are considering the vertical motion
So
[tex]6 = 2 t + \frac{1}{2} * 9.8* t^2[/tex]
[tex]6 = 2 t + 4.9 t^2[/tex]
Solving using quadratic formula w have that
[tex]t = 0.921 \ s[/tex]
Generally the distance of the cork from the balloon is mathematically represented as
[tex]d = v_x * t[/tex]
[tex]d = 5 * 0.921 [/tex]
[tex]d = 4.605 \ m [/tex]
