Questions: A rock is being thrown horizontally at a speed of 12 m / s off the top of an 30 m tall building. How far does the rock land from the base of the building? Please express your answer in three significant figures.

Solution Steps

To find the time it takes for the rock to hit the ground, we use the vertical motion equation under gravity. The initial vertical velocity is zero since the rock is thrown horizontally.

The equation for the vertical displacement is: \[ y = \frac{1}{2} g t^2 \]

Given:

• \( y = 30 \, \text{m} \)
• \( g = 9.81 \, \text{m/s}^2 \)

Solving for \( t \): \[ 30 = \frac{1}{2} \cdot 9.81 \cdot t^2 \] \[ 30 = 4.905 \cdot t^2 \] \[ t^2 = \frac{30}{4.905} \] \[ t^2 \approx 6.115 \] \[ t \approx \sqrt{6.115} \] \[ t \approx 2.473 \, \text{s} \]

The horizontal distance \( x \) can be found using the horizontal velocity and the time of flight.

Given:

• Horizontal velocity \( v_x = 12 \, \text{m/s} \)
• Time of flight \( t \approx 2.473 \, \text{s} \)

The horizontal distance is: \[ x = v_x \cdot t \] \[ x = 12 \cdot 2.473 \] \[ x \approx 29.676 \, \text{m} \]

Final Answer