Please have a hand-written step by step if possible. A particle of mass m = 0.5

Question

Please have a hand-written step by step if possible.

A particle of mass m = 0.5 kg is shot from point P with an initial speed v_0 at an angle of Theta to the horizontal, as shown in the figure. The horizontal component of the initial velocity vector is 30 m/s, and the particle rises to a maximum height of 20 m above point P. Apply what you have learned about conservation of mechanical energy to determine. the vertical component of the initial velocity vector, (20 m/s) the work done by the gravitational force on the particle during its motion from point P to point B, and (300 J) the horizontal, and vertical components of the velocity vector when the particle reaches point B. (v_0x = 30 m/s & v_0y = -40 m/s)

Explanation / Answer

As given in question

Vo*cos(theta) = 30

So,

{Vo * cos(theta)}^2 = 900 ....................(1)

[{ Vo * sin(theta) }^2]/(2 * g) = 20

{Vo*sin(theta) }^2 = 400 .....................(2)

By Adding (1) and (2)

We have

Vo = 36.05551275 m/s

And we also have

Vo * cos (theta) = 30

So,

theta = 33.69006753 degree

a)

So

Vertical component of initial velocity = Vo*sin(theta) = 20m/s

b)

Due to height 60m it gives complex roots so time t is impossible

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