Question

A large building has an inclined roof. The length of the roof is 54.0 m and the angle of the roof is 17.0° below horizontal. A worker on the roof goes of a hammer from the peak of the roof. Starting from rest, it slides down the entire length of the roof with a constant acceleration of 2.87 m/s2. After leaving the edge of the roof, it falls a vertical distance of 46.5 m before hitting the ground.
a. How much time does it take the hammer to fall from the edge of the roof to the ground?
b. How far horizontally does the hammer travel from the edge of the roof until it hits the ground?

Answer 1

Answer:

a) $t=2.6\ s$

b) $s=43.7747\ m$

Explanation:

Given:

length of inclined roof, $l=54\ m$

inclination of roof below horizontal, $\theta=17^{\circ}C$

acceleration of hammer on the roof, $a_r=2.87\ m.s^{-2}$

height from the lower edge of the roof, $h=46.5\ m$

Now, we find the final velocity when leaving the edge of the roof:

Using the equation of motion:

$v^2=u^2+2.a_r.l$

$v^2=0^2+2\times 2.87\times 54$

$v=17.6057\ m.s^{-1}$

The direction of this velocity is 17° below the horizontal.

∴Vertical component of velocity:

$v_y=v.sin\ \theta$

$v_y=17.6057\times sin\ 17^{\circ}$

$v_y=5.1474\ m.s^{-1}$

a.

So, the time taken to fall on the ground:

$h=ut+\frac{1}{2} g.t^2$

here:

initial velocity, $u=v_y=5.1474\ m.s^{-1}$

putting respective values

$46.5=5.1474\times t+0.5\times 9.8\times t^2$

$t=2.6\ s$

b.

Horizontal component of velocity, $v_x=v.cos\ \theta=17.6057\ cos\ 17^{\circ}=16.8364\ m.s^{-1}$

Since there is no air resistance so the horizontal velocity component remains constant.

∴Horizontal distance from the edge of the roof where the hammer falls is given by:

$s=v_x.t$

$s=16.8364\times 2.6$

$s=43.7747\ m$