A grasshopper leaps into the air from the edge of a vertical cliff. Assume that

1 answer:

7 0

Maximum height reached h = 7.15cm = 0.0715 m The horizontal distance b = 1.04 m Angle of jump = 49 degrees. We have v^2 = 2gh which here is (vsin theta)^2 = 2gh
v^2 = (2 x 9.81 x 0.0715) / sin^2 49 = 1.402 / -0.95^2 = 1.55 => v = 1.24 m/s
Time taken for flight = b / v Cos theta = 1.04 / (1.24 x cos49)
= 1.04 / 0.8135 =1.27 s
Time to reach maximum height = vsin theta / g = (1.24 x 0.95) / 9.8 = 0.12 s
Now time taken for top to ground = 1.27 - 0.12 = 1.15 s
So the total height H =1/2 gt = 1/2 x 9.8 x 1.15 = 5.635.
Now the height of the cliff = 5.635 - 0.0715 = 5.56 m

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The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

$x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg$

Since the frictional force $f_k$ is equal to $f_k=\mu_k N=\mu_k mg$ , then we have that the acceleration is:

$a=\frac{\mu_k mg}{m}=\mu_k g$

Now, from the definition of acceleration we get:

$a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}$

And, as the final velocity is zero because the box gets to a stop, we have:

$t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}$

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

$t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s$