Question

From the edge of a cliff, a 0.46 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum upward displacement from the launch point is +150 m. What are the (a) horizontal and (b) vertical components of its launch velocity? (c) At the instant the vertical component of its velocity is 65 m/s, what is its vertical displacement from the launch point? Take up to be the positive direction.

Answer 1

Answer:

Explanation:

Given

mass of Projectile(m)=0.46 kg

Initial Kinetic Energy=1430 J

Maximum upward displacement from Launch point=150 m

$K.E.=\frac{mv^2}{2}$

$1430\times 2=0.46\times v^2$

v=78.85 m/s

and $H_{max}=\frac{v^2\sin ^2\theta }{2g}$

$150=\frac{(78.85)^2\sin ^2\theta }{2\times 9.8}$

$\sin \theta =0.687$

$\theta =43.39^{\circ}$

initial horizontal velocity$(v_x)=v\cos \theta$

$v_x=78.84\times \cos (43.39)=57.29 m/s$

Initial vertical velocity$(v_y)=v\sin \theta$

$v_y=78.85\times \sin (43.39)$

$v_y=54.16 m/s$

(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

$v^2-u^2=2 a s$

$65^2-54.16^2=2\times 9.8\times s$

$s=65.90 m$