Answer:
Part a)
[tex]H = 2.44 m[/tex]
Part b)
[tex]H_1 + H_2 = 1.95 m[/tex]
Explanation:
Part a)
As we know that there is no friction on the inclined plane
So we can use energy conservation for both planes
For longer plane we will have
[tex]mgH = \frac{1}{2}mv^2[/tex]
[tex]H = \frac{v^2}{2g}[/tex]
[tex]H = \frac{6.92^2}{2(9.81)}[/tex]
[tex]H = 2.44 m[/tex]
Part b)
On shorter plane the speed while it leave the plane at height H1 is given as
[tex]\frac{1}{2}m(v_f^2 - v_i^2) = -mgH_1[/tex]
[tex]\frac{1}{2}m(v_f^2 - 6.92^2) = -(9.81)(1.25)[/tex]
[tex]v_f = 4.83 m/s[/tex]
now the maximum height of projectile is given as
[tex]H_2 = \frac{v_f^2 sin^2\theta}{2g}[/tex]
[tex]H_2 = \frac{4.83^2sin^2(50)}{2(9.81)}[/tex]
[tex]H_2 = 0.70 m[/tex]
so we have
[tex]H_1 + H_2 = 1.25 + 0.7[/tex]
[tex]H_1 + H_2 = 1.95 m[/tex]
