# Kinematics - Two particles release from same height ...

Good afternoon (from Europe) !

The given solution of mentioned problem is wrong.

There must be some initial velocity of particle "A" moving downwards along slope.

Particle "\"B\"": t_B=√(2h/g)

Particle "\"A\" ": L=v_(A,i) t+a t^2/2=v_(A,i) √(2h/g)+gsinα (√(2h/g))^2/2=v_(A,i) √(2h/g)+h^2/L

v_(A,i)=L(1-h^2/L^2 ) √(g/2h)=20*(1-〖12〗^2/〖20〗^2 ) √(9.81/(2*12))=8.18 m/s

The result meets our expectation.

Best regards !

congestus

## Link to article: http://www

Link to article: http://www.mathalino.com/node/2976

Time for A and B to reach the base:

$t = \sqrt{24/g} = 1.564 ~ \text{sec}$

From your solution$20 = 8.18t + \frac{1}{2}(\frac{12}{20}g)t^2$

$t = 1.564 ~ \text{sec}$ (

okay)Components of 8.18 m/s

$v_{ox} = \frac{4}{5}(8.18) = 6.544 ~ \text{m/sec}$

$v_{oy} = \frac{3}{5}(8.18) = 4.908 ~ \text{m/sec}$

Check the time in horizontal motion (a = 0 and s = 16 m)

$s = v_ot + \frac{1}{2}at^2$

$16 = 6.544t + 0$

$t = 2.445 ~ \text{sec}$ (

not okay)Check the time in vertical motion (a = g and s = 12 m)

$s = v_ot + \frac{1}{2}at^2$

$12 = 4.908t + \frac{1}{2}gt^2$

$t = 1.142 ~ \text{sec}$ (

not okay)Solution from the post (link: http://www.mathalino.com/node/2976)Let 2 be the subscript indicating the base of the incline

$v = v_o + at$

$v_{2y} = v_{oy} + at$

$v_{2y} = 0 + g\sqrt{24/g}$

$v_{2y} = 15.344 ~ \text{m/sec}$

Inclined Velocity

$\dfrac{v_2}{15.344} = \dfrac{20}{12}$

$v_2 = 25.573 ~ \text{m/sec}$

Horizontal Velocity

$\dfrac{v_{2x}}{15.344} = \dfrac{16}{12}$

$v_{2x} = 20.459 ~ \text{m/sec}$

Check the Inclined Motion

$v^2 = {v_o}^2 + 2as$

${v_2}^2 = {v_o}^2 + 2aL$

$25.573^2 = 0^2 + 2a(20)$

$a = 16.349 ~ \text{m/sec}^2$

$v = v_o + at$

$v_2 = v_o + at$

$25.573 = 0 + 16.349t$

$t = 1.564 ~ \text{sec}$ (

okay)Check the Horizontal Motion

$v^2 = {v_o}^2 + 2as$

${v_{2x}}^2 = {v_{ox}}^2 + 2a_x L_x$

$20.459^2 = 0^2 + 2a_x (16)$

$a_x = 13.08 ~ \text{m/sec}^2$

$v = v_o + at$

$v_{2x} = v_{ox} + a_x t$

$20.459 = 0 + 13.08t$

$t = 1.564 ~ \text{sec}$ (

okay)Additional checking: the a

_{y}must be equal to g${a_x}^2 + {a_y}^2 = a^2$

$13.08^2 + {a_y}^2 = 16.349^2$

$a_y = 9.81 ~ \text{m/sec}^2$ (

okay)PS: The solution you presented will only work if the path is projectile, it won't work to inclined straight line.