Question

level 4 - multi - step application

1. a carpenter tosses a tool horizontally from a 7.8 m high roof at a speed of 6.5 m/s.

a. how long does it take to reach the ground?
b. what is its final vertical velocity just before hitting the ground?

Explanation:

Step1: Analyze vertical motion

The vertical motion of the tool is a free - fall motion. The initial vertical velocity \(v_{0y}=0\ m/s\), the vertical displacement \(y - y_0=- 7.8\ m\) (taking downwards as negative), and the acceleration \(a = g=-9.8\ m/s^{2}\). Use the equation \(y - y_0=v_{0y}t+\frac{1}{2}at^{2}\). Substituting the values, we get \(-7.8 = 0\times t+\frac{1}{2}\times(-9.8)t^{2}\).

Step2: Solve for time \(t\)

From \(-7.8=\frac{-9.8}{2}t^{2}\), we can rewrite it as \(t^{2}=\frac{7.8\times2}{9.8}\). Then \(t=\sqrt{\frac{15.6}{9.8}}\approx1.26\ s\).

Step3: Find final vertical velocity

Use the equation \(v_y = v_{0y}+at\). Since \(v_{0y} = 0\ m/s\), \(a=-9.8\ m/s^{2}\) and \(t\approx1.26\ s\), then \(v_y=0+(-9.8)\times1.26=- 12.35\ m/s\) (the negative sign indicates the direction is downwards).

Answer:

a. The time it takes to reach the ground is approximately \(1.26\ s\).
b. The final vertical velocity just before hitting the ground is approximately \(-12.35\ m/s\).