Answer to Question #137406 in Molecular Physics | Thermodynamics for Max

Question

Two bodies were thrown simultaneously from the same point: one, straight up, and the other, at an angle of θ = 60° to the horizontal. The initial velocity of each body is equal to v0 = 25 m/s. Neglecting the air drag, find the distance between the bodies t = 1.70 s later.

Answer 1

$v_0 = 25m/s$

$t = 1.70s$

Let the body thrown straight up be 1, and the other one be 2

Note that;

$a_y = -g = -\,10m/s^2$

$ax = 0\,m/s^2$

For body 1

$\theta$₁ = $90\text{\textdegree}$ (i.e vertical)

$\begin{aligned} U_{y1} = v_0sinθ \quad &; \quad U_{x1} = v_0cosθ\\ U_{y1} = 25sin90 \enspace &; \quad U_{x1} = 25cos90\\ U_{y1} = 25m/s \quad & ;\quad U_{x1} = 0m/s \end{aligned}$

$\begin{alignedat}{2} &x_1 = U_{x1}t\, &+& \,0.5a_xt^2\\ &x_1 = 0(1.70)\, &+&\, 0.5(0)(1.70)^2\\ &x_1 = 0 + 0\\ ∴& x_1 = 0m \end{alignedat}$

$\begin{alignedat}{2} &y_1 = U_{y1}t \,&+& \,0.5a_yt^2\\ &y_1 = (25)(1.70) \,&+&\, 0.5(-10)(1.70)^2\\ &y_1 = 42.5\, &-&\, 14.45\\ ∴& y_1 = 28.05m \end{alignedat}$

For body 2

$\theta$₂ = $60\text{\textdegree}$

$\begin{aligned} U_{y2} = v_0sinθ \quad \,\,\, &; \quad U_{x2} = v_0cosθ\\ U_{y2}= 25sin60 \quad & ; \quad U_{x2} = 25cos60\\ U_{y2}= 21.65m/s\,\, & ; \quad U_{x2} = 12.5m/s \end{aligned}$

$\begin{alignedat}{2} &x_2 = U_{x2}t \, &+&\, 0.5a_xt^2\\ &x_2 = 12.5(1.70) \,&+& \,0.5(0)(1.70)^2\\ &x_2 = 21.25 &+&\, 0\\ ∴ &x_2 = 21.25m \end{alignedat}$

$\begin{alignedat}{2} &y_2 = U_{y2}t \,&+&\, 0.5a_yt^2\\ &y_2 = (21.65)(1.70)\, &+&\, 0.5(-10)(1.70)^2\\ &y_2 = 36.81 &-& \,14.45\\ ∴&y_2= 22.36m \end{alignedat}$

from, distance between 2 points = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

distance between the 2 bodies = $\sqrt{(21.25-0)^2+(22.36-28.05)^2}$

= $\sqrt{(21.25)^2+(-5.69)^2}$

= $\sqrt{451.56+32.38}$

= $\sqrt{483.94}$

= $22m$

$\therefore$ the distance between the two bodies is 22m