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

Step-by-step explanation:

Given

mass of Projectile(m)=0.46 kg

Initial Kinetic Energy=1430 J

Maximum upward displacement from Launch point=150 m

`K.E.=(mv^2)/(2)`

`1430* 2=0.46* v^2`

v=78.85 m/s

and
`H_(max)=(v^2\sin ^2\theta )/(2g)`

`150=((78.85)^2\sin ^2\theta )/(2* 9.8)`

`\sin \theta =0.687`

`\theta =43.39^(\circ)`

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

`v_x=78.84* \cos (43.39)=57.29 m/s`

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

`v_y=78.85* \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* 9.8* s`

`s=65.90 m`