Question

A baseball player hits a baseball at an angle of 30 degrees above the horizontal. The baseball’s initial speed is 35 m/s.

Answer 1

The ball's velocity at 2 sec is 30.4 m/s at $-4.0^{\circ}$ (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

$v_x = u cos \theta$

where

u = 35 m/s is the initial speed of the ball

$\theta=30^{\circ}$ is the angle of projection

Substituting,

$v_x = (35)(cos 30)=30.3 m/s$

- The vertical velocity is changing, and its value is given by

$v_y = u_y + at$

where

$u_y = u sin \theta = (35)(sin 30)=17.5 m/s$ is the initial vertical velocity

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it points down)

t is the time

At t = 2 s,

$v_y = 17.5 + (-9.8)(2)=-2.1 m/s$

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:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{30.3^2+(-2.1)^2}=30.4 m/s$

While the direction is given by

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-2.1}{30.3})=-4.0^{\circ}$

Learn more about projectile motion:

brainly.com/question/8751410

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