Respuesta :
Answer:
Explanation:
Given
mass of Projectile(m)=0.46 kg
Initial Kinetic Energy=1430 J
Maximum upward displacement from Launch point=150 m
[tex]K.E.=\frac{mv^2}{2}[/tex]
[tex]1430\times 2=0.46\times v^2[/tex]
v=78.85 m/s
and [tex]H_{max}=\frac{v^2\sin ^2\theta }{2g}[/tex]
[tex]150=\frac{(78.85)^2\sin ^2\theta }{2\times 9.8}[/tex]
[tex]\sin \theta =0.687[/tex]
[tex]\theta =43.39^{\circ}[/tex]
initial horizontal velocity[tex](v_x)=v\cos \theta [/tex]
[tex]v_x=78.84\times \cos (43.39)=57.29 m/s[/tex]
Initial vertical velocity[tex](v_y)=v\sin \theta [/tex]
[tex]v_y=78.85\times \sin (43.39)[/tex]
[tex]v_y=54.16 m/s[/tex]
(c)vertical velocity at any instant=65 m/s
Since initial vertical velocity is 54.16 m/s
so 65 m/s will be acquired when projectile started falling below cliff
[tex]v^2-u^2=2 a s[/tex]
[tex]65^2-54.16^2=2\times 9.8\times s[/tex]
[tex]s=65.90 m[/tex]
