Question

Particle mass of 0.5 kg. The particle has a initial velocity with horizontal component of 30 m/s. The particle rises to a max height of 20m above p.
partical is on a cliff 60 m high. Find velocity in the y direction (vertical). Find horizontal and veryical components when the particle reaches b at the bottom.

Answer 1

Answer on Question #50339, Physics, Mechanics | Kinematics | Dynamics

Task:

Particle mass of $0.5\mathrm{kg}$ . The particle has a initial velocity with horizontal component of $30\mathrm{m/s}$ . The particle rises to a max height of $20\mathrm{m}$ above $\mathfrak{p}$ .

partical is on a cliff $60\mathrm{m}$ high. Find velocity in the y direction (vertical). Find horizontal and vertical components when the particle reaches b at the bottom

Answer:

Equations of motion:

$x(t) = V_{x0}t + x_0$

$y(t) = -gt^{2} / 2 + V_{y0}t + y_{0}$

Initial conditions: $x_0 = 0$ $0 = 0$ $V_{x0} = 30$

Thus, $(t) = V_{x0}t$ $y(t) = -gt^{2} / 2 + V_{y0}t$

1) velocity in the y direction (vertical) $V_{oy}$ :

the maximum height of the particle is a value of $y(t)$ at time, when $dy(t) / dt = 0$ .

$dy(y) / dt = -gt + V_{y0} = 0$ , $t = V_{y0} / g$ , $y(V_{y0} / g) = a$ max height of $20m = -g(V_{y0} / g)^2 / 2 + V_{y0}(V_{y0} / g) =$

$V_{y0}^{2} / 2g$ , thus $V_{oy} = (2gh)^{1/2} = (2*10*20)^{1/2} = 20m/s$ .

2) horizontal and vertical components when the particle reaches $b$ at the bottom:

$V_{y} = V_{oy} + gt$ , $t = (V_{oy} + (V_{oy}^{2} + 2gh)^{1/2}) / g = 6s$ ; so $V_{y} = 80m/s$ .

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