Question

An apple is pushed across a table with a velocity of 3.74 m/s rolls too far and falls 0.89

Answer 1

The apple lands 1.61 m from the base of the table

Explanation:

The motion of the apple here 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  

First we analyze the vertical motion, to find the time of flight of the apple. We can do it by using the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where, choosing downward as positive direction,  we have:

s = 0.89 m is the vertical displacement of the apple

$u_y=0$ is the initial vertical velocity  (since it is thrown horizontally)

t is the time of the fall

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t,  we find

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(0.89)}{9.8}}=0.43 s$

So the apple takes 0.43 s to fall to the ground.

Now we can analyze the horizontal motion: the apple moves horizontally with a constant velocity of

$v_x = 3.74 m/s$  (its initial velocity)

Therefore, the distance it covers in a time t is given by

$d=v_x t$

and by substituting t = 0.43 s, we find the final distance of the apple from the base of the table:

$d=(3.74)(0.43)=1.61 m$

Learn more about projectile motion:

brainly.com/question/8751410

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