Question

question 1 (1 point)
an object 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. how long does it take for the object to return to where it is thrown?
6.0 s
8.0 s
10 s
12 s

Explanation:

Step1: Analyze the motion

The object is in free - fall motion. The initial velocity \(v_0 = 15\ m/s\) and the acceleration \(a=- 2.5\ m/s^2\) (negative because it acts against the direction of motion). When the object returns to the starting point, the displacement \(y - y_0=0\). We use the kinematic equation \(y - y_0=v_0t+\frac{1}{2}at^2\).
Substituting \(y - y_0 = 0\), \(v_0 = 15\ m/s\) and \(a=-2.5\ m/s^2\) into the equation \(0 = 15t-\frac{1}{2}\times2.5t^2\).
Factor out \(t\): \(t(15 - 1.25t)=0\).
One solution is \(t = 0\) (corresponds to the initial time when the object is thrown).
For the non - zero solution, we solve \(15-1.25t = 0\).

Step2: Solve for \(t\)

$$
LATEXBLOCK0
$$

Another way:
The time to reach the maximum height is found using \(v = v_0+at\). At the maximum height, \(v = 0\).

$$
LATEXBLOCK1
$$

The time to come down from the maximum height is the same as the time to go up (because the motion is symmetric in the absence of air resistance). So \(t_{total}=t_{up}+t_{down}\). Since \(t_{up}=t_{down}\), \(t_{total}=2t_{up}\)

$$
LATEXBLOCK2
$$

Answer:

\(12\ s\)