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A cat chase a mouse across a 1.0 m high table. The mouse steps out of the way, and the car slides off the table and strikes the floor 2.2m from the edge of the table. When the car slid off the table what was it’s speed?

Answer is 4.9 m/s how do we get 4.9 m/s?

Respuesta :

zachmiller

The cat fell 1.0 m from the ground.

Using the formula

[tex]h = v_{oy}t - \frac{gt^2}{2}[/tex]

Here, [tex]v_{oy} = 0, \; h = 1.0 \; m.[/tex]

Solving for t, the time it spent in the air is

[tex]t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.0 \; m)}{9.8 \; m/s^2}} = 0.451753951 \; s[/tex]

The cat does not accelerate along the horizontal, so it has constant horizontal velocity. Since it strikes the floor 2.2 m from the table, then

[tex] v_x = \Delta x/t = \frac{2.2\; m}{0.451753951 s} = 4.869907597 \; m/s \Rightarrow 4.9 \; m/s [/tex]


snehashish65

The car slid off the table at a speed of 4.90 m/s.

Given data:

The height of table is, [tex]h = 1.0 \;\rm m[/tex].

Horizontal distance slide by the car is, [tex]d=2.2 \;\rm m[/tex].

Apply the second kinematic equation of motion as,

[tex]h= ut+\dfrac{1}{2}gt^{2}[/tex]

Here, u is the initial speed of car, g is the gravitational acceleration and t is the time.

[tex]1.0= 0 \times t+\dfrac{1}{2} \times 9.8t^{2}\\1.0= \dfrac{1}{2} \times 9.8t^{2}\\t=\sqrt{\dfrac{2}{9.8}} \\t=0.451 \;\rm s[/tex]

Now, the speed of car is calculated as,

[tex]v=\dfrac{d}{t}\\v=\dfrac{2.2}{0.451}\\v \approx 4.90 \;\rm m/s[/tex]

Thus, the car slid off the table at a speed of 4.90 m/s.

Learn more about kinematic equations of motion here:

https://brainly.com/question/14355103?referrer=searchResults

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