Answer:15.20 m
Explanation:
Given
initial velocity [tex](v_i)=15 m/s[/tex]
inclined length=10 m
[tex]\mu _k=0.435 [/tex]
inclination [tex]\theta =43.5^{\circ}[/tex]
[tex]F_{net}=f_r+mg\sin \theta [/tex]
[tex]a_{net}=\mu _kg\cos \theta +g\sin \theta [/tex]
[tex]a_{net}=0.435\times 9.8\times \cos 43.5+9.8\times \sin 43.5[/tex]
[tex]a_{net}=3.09+6.74=9.83 m/s^2[/tex]
[tex]v^2-u^2=2as[/tex]
[tex]v^2=15^2-2\times (9.83)10[/tex]
[tex]v=5.31 m/s[/tex]
So Particle launches with a speed of 5.31 m/s at an angle of \theta [tex]=43.5[/tex]
[tex]h_{max}=\frac{u^2\sin^2\theta }{2g}[/tex]
[tex]h_{max}=\frac{(5.31)^2\sin ^2(43.5)}{2\times 9.81}[/tex]
[tex]h_{max}=0.679[/tex]
Total height raised is [tex]0.679+\frac{10}{\sin 43.5} =15.20 m[/tex]
