Question

A worker on the roof of a house drops his hammer, which slides down the roof at a constant speed of 4 m/s. The roof makes an angle of 30 degree with the horizontal, and its lowest point is 10 m above the ground. What is the horizontal distance traveled by the hammer after it leaves the roof of the house and before it hits the ground? Use g 10 m/s2.

Answer 1

Answer:

4.25m

Explanation:

The horizontal and vertical component of the velocity as it leaves the roof is

$v_v = vsin30^0 = 4*0.5 = 2 m/s$

$v_h = vcos30^0 = 4*0.866 = 3.64 m/s$

Neglect air resistance, then gravitational acceleration g = 10m/s2 is the only thing that affect the vertical motion of the hammer. We can use the following equation of motion to solve for the time t of dropping

$h = v_vt + gt^2/2$

where h = 10m is the distance of dropping before the hammer reaches the ground

$10 = 2t + 5t^2$

$5t^2 + 2t - 10 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{-2\pm \sqrt{(2)^2 - 4*(5)*(-10)}}{2*(5)}$

$t= \frac{-2\pm14.28}{10}$

t = 1.23 or t = -1.63

Since t can only be positive we will pick t = 1.23

t is also the time that it takes to travel horizontally at the constant rate of 3.64 m/s

So the horizontal distance traveled by the hammer is

3.64*1.23 = 4.25 m