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quester [9]
2 years ago
15

A lab cart flies off the end of a track on a lab table with a horizontal velocity of 2.08 m/s. If it lands 0.96 meters from the

edge of the table, how tall is the table?
A)0.5 m
B)6.0 m
C)1.2 m
D)2.2 m
​
Physics
1 answer:
lana [24]2 years ago
3 0

Answer:

Option C. 1.2 m

Explanation:

The following data were obtained from the question:

horizontal velocity (u) = 2.08 m/s

Horizontal distance (s) = 0.96 m

Height (h) of the table =?

Next, we shall determine the time taken for the lab cart to get to the ground. This can be obtained as follow:

horizontal velocity (u) = 2.08 m/s

Horizontal distance (s) = 0.96 m

Time (t) =?

s = ut

0.96 = 2.08 × t

Divide both side by 2.08

t = 0.96 / 2.08

t = 0.5 s

Finally, we shall determine the height of the table as illustrated below:

Time (t) = 0.5 s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) of the table =?

h = ½gt²

h = ½× 9.8 × 0.5²

h = 4.9 × 0.25

h = 1.2 m

Thus, the height of the table is 1.2 m

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How do you solve this problem?
Katarina [22]

The particle has acceleration vector

\vec a=\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\jmath

We're told that it starts off at the origin, so that its position vector at t=0 is

\vec r_0=\vec0

and that it has an initial velocity of 12 m/s in the positive x direction, or equivalently its initial velocity vector is

\vec v_0=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\,\vec\imath

To find the velocity vector for the particle at time t, we integrate the acceleration vector:

\vec v=\vec v_0+\displaystyle\int_0^t\vec a\,\mathrm d\tau

\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath

\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath

Then we integrate this to find the position vector at time t:

\vec r=\vec r_0+\displaystyle\int_0^t\vec v\,\mathrm d\tau

\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath

\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath

Solve for the time when the y coordinate is 18 m:

18\,\mathrm m=\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\implies t=3.0\,\mathrm s

At this point, the x coordinate is

\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)(3.0\,\mathrm s)+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3.0\,\mathrm s)^2=27\,\mathrm m

so the answer is C.

7 0
3 years ago
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