Question

A plate is pushed horizontally over a 80 cm high table. If it lands a point 120 cm away from the table
A) What is the initial velocity of the plat ?
B)At what velocity does theolate strike the floor ?

Answer 1

A) The initial velocity of the plate is 3.0 m/s

B) The final velocity of the plate is 4.9 m/s

Explanation:

A)

The motion of the plate is an example of projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall) due to the presence of gravity

We analyze the vertical motion first to find the time of flight of the plate. Since the motion is a uniformly accelerated motion, we use the suvat equation:

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

where

s = 80 cm = 0.80 m is the vertical displacement

u = 0 is the initial vertical velocity

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

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(0.80)}{9.8}}=0.40 s$

Now we now that the horizontal distance covered by the plate is

d = 120 cm = 1.20 m

So, the horizontal velocity (which is constant) is

$v_x =\frac{d}{t}= \frac{1.20}{0.40}=3.0 m/s$

And this was the initial velocity of the plate, since it was thrown horizontally.

B)

We have to find the horizontal and the vertical components of the velocity of the plate when it strikes the floor.

The horizontal velocity is constant during the motion, therefore it is equal to

$v_x = 3.0 m/s$

While the vertical velocity follows the equation:

$v_y = u_y + at$

where

$u_y = 0$ is the initial vertical velocity

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

t = 0.40 s is the time of flight

Substituting,

$v_y = 0 +(9.8)(0.4)=3.9 m/s$

And therefore, the velocity of the plate when it strikes the floor is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(3.0)^2+(3.9)^2}=4.9 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

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