Question

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/s2 and there is no atmosphere. How long does it take for the instrument to return to where it was thrown?

Answer 1

Using the kinematic equation for free fall, the time it takes for the instrument to reach the point of zero velocity on planet X is 6 seconds. Since the descent takes an equal amount of time as the ascent, the total round trip time is 12 seconds.

Step-by-step explanation:

To determine the time it takes for the instrument to return to its original position, we can use the kinematic equation for free fall motion under uniform acceleration, which is given by:

v = u + at

Where:

• v is the final velocity (0 m/s at the highest point)
• u is the initial velocity (15 m/s)
• a is the acceleration due to gravity (-2.5 m/s^2; negative because it's opposite the direction of initial velocity)
• t is the time

Rearranging the equation to solve for t:

t = (v - u) / a

The time it takes to reach the highest point is:

t = (0 m/s - 15 m/s) / (-2.5 m/s^2) = 6 seconds

To find the total time for the round trip, we need to double this time because the descent will take the same amount of time as the ascent:

Total time = ascent time + descent time = 6 s + 6 s = 12 seconds.

Answer 2

Answer: 12 s

Step-by-step explanation:

The situation described here is parabolic movement. However, as we are told the instrument is thrown upward from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

`y-y_(o)=V_(o).t-(1)/(2)g.t^(2)` (1)

Where:

`y` is the instrument's final position

`y_(o)=0` is the instrument's initial position

`V_(o)=15m/s` is the instrument's initial velocity

`t` is the time the parabolic movement lasts

`g=2.5(m)/(s^(2))` is the acceleration due to gravity at the surface of planet X.

As we know
`y_(o)=0` and
`y=0` when the object hits the ground, equation (1) is rewritten as:

`0=V_(o).t-(1)/(2)g.t^(2)` (2)

Finding
`t`:

`0=t(V_(o)-(1)/(2)g.t^(2))` (3)

`t=(2V_(o))/(g)` (4)

`t=(2(15m/s))/(2.5(m)/(s^(2)))` (5)

Finally:

`t=12s`