The ball's velocity at 2 sec is 30.4 m/s at [tex]-4.0^{\circ}[/tex] (below the horizontal)
Explanation:
The motion of the ball is a projectile motion, therefore it consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction
This means that:
- The horizontal velocity of the ball is constant, and it is given by
[tex]v_x = u cos \theta[/tex]
where
u = 35 m/s is the initial speed of the ball
[tex]\theta=30^{\circ}[/tex] is the angle of projection
Substituting,
[tex]v_x = (35)(cos 30)=30.3 m/s[/tex]
- The vertical velocity is changing, and its value is given by
[tex]v_y = u_y + at[/tex]
where
[tex]u_y = u sin \theta = (35)(sin 30)=17.5 m/s[/tex] is the initial vertical velocity
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity (negative because it points down)
t is the time
At t = 2 s,
[tex]v_y = 17.5 + (-9.8)(2)=-2.1 m/s[/tex]
where the negative sign means that at t = 2 s, the vertical velocity points down.
Now we can find the magnitude of the ball's velocity at 2 sec:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{30.3^2+(-2.1)^2}=30.4 m/s[/tex]
While the direction is given by
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-2.1}{30.3})=-4.0^{\circ}[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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