Question

A worker kicks a flat object lying on a roof. The object slides up the incline 10.0 m to the apex of the roof, and flies off the roof as a projectile. What maximum height (in m) does the object attain? Assume air resistance is negligible, vi = 15.0 m/s, μk = 0.435, and that the roof makes an angle of θ = 43.5° with the horizontal. (Assume the worker is standing at y = 0 when the object is kicked.

Answer 1

Answer:15.20 m

Explanation:

Given

initial velocity $(v_i)=15 m/s$

inclined length=10 m

$\mu _k=0.435$

inclination $\theta =43.5^{\circ}$

$F_{net}=f_r+mg\sin \theta$

$a_{net}=\mu _kg\cos \theta +g\sin \theta$

$a_{net}=0.435\times 9.8\times \cos 43.5+9.8\times \sin 43.5$

$a_{net}=3.09+6.74=9.83 m/s^2$

$v^2-u^2=2as$

$v^2=15^2-2\times (9.83)10$

$v=5.31 m/s$

So Particle launches with a speed of 5.31 m/s at an angle of \theta $=43.5$

$h_{max}=\frac{u^2\sin^2\theta }{2g}$

$h_{max}=\frac{(5.31)^2\sin ^2(43.5)}{2\times 9.81}$

$h_{max}=0.679$

Total height raised is $0.679+\frac{10}{\sin 43.5} =15.20 m$