On Planet X, you drop a 25-kg stone from rest and measure its speed at various t

Question

On Planet X, you drop a 25-kg stone from rest and measure its speed at various times. Then you use the data you obtained to construct a graph of its speed v as a function of time (t). From the information in the graph, answer the following questions:

a,b and c from the picture below

5. On Planet X, you drop a 25-kg stone from rest and measure its speed at various times. Then you use the data you obtained to construct a graph of its speed v as a function of time (t). From the information in the graph, answer the following questions 1 of 1 (a) What is g on Planet X? (b) An astronaut drops a piece of equipment from rest out of the landing module, 3.5 m above the surface of Planet X. How long will it take this equipment to reach the ground, and how fast will it be moving when it gets there? (c) How fast would an astronaut have to project an object straight upward to reach a height of 18.0 m above the release point, and how long would it take to reach that height? v (m/s) 30 20 10

Explanation / Answer

Given that,

mass of the stone = m = 25 kg

(a) We need to find g.

We know that the slope of velocity vs time graph is acceleration. So,

g = a(y) = v(y) / t = 30 / 2 = 15 m/s2

Hence, g = 15 m/s2.

(b) h = 3.5 m, Let v be the velocity with which it is moving.

From third equation of motion we know that,

h = y0 + u(y) x t + 1/2 a(y) t2 ( u(y) = 0 and y0 = 0)

solving for t, we get

t = sqrt [2 x h / a(y)] = sqrt ( 2 x 3.5 / 15) = sqrt (0.467) = 0.68 sec

so we get, t = 0.68 s

from first equation of motion,

v = u + a(y) t = 0 + 15 x 0.68 = 10.2 m/s

Hence, v = 10.2 m/s

(c) Given, s = 18m. Let t be the time.

We know that, at maximum height, u(y) = 0

from equation of motion,

v(y)2 = u(y)2 + 2 a(y) s

v(y) = sqrt [ 2 a(y) s ] = sqrt ( 2 x 15 x 18 ) = sqrt (540) = 23.2 m/s

Now again from first eqn of motion

v(y) = u(y) + a(y) x t => t =[ v(y) - u(y)] / a(y) = ( 23.2 - 0 ) / 15 = 1.55 sec

Hence, t = 1.55 sec.

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