Question

an object is thrown upward with a speed of 12 m/s on the surface of planet x where the acceleration due to gravity is 1.5 m/s2. what is the maximum height reached by the object?
8.0 m
18 m
48 m
144 m

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation \( v_f^2 = v_i^2 + 2a\Delta y \), where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, \( a \) is the acceleration, and \( \Delta y \) is the displacement (height in this case). At maximum height, the final velocity \( v_f = 0 \, \text{m/s} \). The acceleration \( a \) is equal to \( -g \) (negative because it acts opposite to the initial velocity) where \( g = 1.5 \, \text{m/s}^2 \), and the initial velocity \( v_i = 12 \, \text{m/s} \).

Step2: Rearrange the equation to solve for \( \Delta y \)

Rearranging \( v_f^2 = v_i^2 + 2a\Delta y \) for \( \Delta y \), we get:

$$ \Delta y=\frac{v_f^2 - v_i^2}{2a} $$

Substituting \( v_f = 0 \), \( v_i = 12 \, \text{m/s} \), and \( a=- 1.5 \, \text{m/s}^2 \):

$$ \Delta y=\frac{0-(12)^2}{2\times(- 1.5)} $$

Step3: Calculate the value

First, calculate the numerator: \( 0 - 144=- 144 \)
Then, calculate the denominator: \( 2\times(-1.5) = - 3 \)
Now, divide the numerator by the denominator:

$$ \Delta y=\frac{-144}{-3}=48 \, \text{m} $$

Answer:

48 m (corresponding to the option "48 m")