Answer to Question #245573 in Mechanics | Relativity for Audrey

Question

A carpenter on the roof accidentally drops the nail that hits the ground after 2.5 seconds. How high is the roof? What is the velocity of the nail just before hitting the ground?

Answer 1

Explanations & Calculations

1)

• The nail falls under the constant acceleration: g until it reaches the ground. Therefore, applying $\small S = \small ut+1/2 at^2$, you can calculate the height it travels thus the height of the roof.

$\qquad\qquad \begin{aligned} \small \downarrow h &= \small 0ms^{-1}\times 2.5 s+\frac{1}{2}\times 9.8ms^{-2}\times (2.5s)^2 \\ \small h&=\small 30.6\,m \end{aligned}$

2)

• Applying $\small v =u+at$, you can calculate the velocity with what it hits the floor.

$\qquad\qquad \begin{aligned} \small v&=\small 0ms^{-1}+9.8ms^{-2}\times 2.5s\\ \small v&=\small 24.5\,ms^{-1} \end{aligned}$