The velocity after 2.0 s is 19.64 m/s at 86.5 degrees below the horizontal
Explanation:
The motion of the ball consists of two independent motions:
- Along the horizontal motion, it is a uniform motion with constant horizontal speed of [tex]v_x = 1.2 m/s[/tex]
- Along the vertical motion, it is a free fall motion with constant acceleration of [tex]g=9.8 m/s^2[/tex] towards the ground
The horizontal velocity is therefore constant, and when the marble reaches the ground, it is
[tex]v_x = 1.2 m/s[/tex]
To find the vertical component of the velocity, we use the suvat equation:
[tex]v_y = u_y - gt[/tex]
where
[tex]u_y = 0[/tex] is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex]
t = 2.0 s is the time of flight of the ball
Substituting,
[tex]v_y = 0-(9.8)(2.0)=-19.6 m/s[/tex]
where the negative sign means the direction is downward.
Therefore, the magnitude of the velocity of the marble when it hits the ground is:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{1.2^2+(-19.6)^2}=19.64 m/s[/tex]
And the direction is
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-19.6}{1.2})=-86.5^{\circ}[/tex]
below the horizontal.
Learn more about projectile motion here:
brainly.com/question/8751410
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