Answer to Question #108236 in Mechanics | Relativity for havefun7741

Question

[img]https://upload.cc/i1/2020/04/06/gJcTdA.jpg[/img]



question is in the picture, also d8=4

Answer 1

By the condition of the problem

$V_0=5m/s$ $a=1.5 m/s^2$

During the fall of the ball from a height $h$ The car will move to a point $x$

From the equation $h=\frac{g \cdot t^2}{2}$

Determine the time

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2 \cdot 20 \cdot \sin(60^0)}{9.81}}=1.879 s$

We write the value of the coordinate $x$ for ball and car

 $x=u_x \cdot t$

$x=x_0+V_0 \cdot t +\frac{a \cdot t^2}{2}$

we equate these equations

$u_x \cdot t=x_0+V_0 \cdot t +\frac{a \cdot t^2}{2}$

a)Where will we write (initial ball speed)

$u_x =\frac{x_0}{t}+V_0 +\frac{a \cdot t}{2}$

$u_x =\frac{20 \cdot \cos{60^0+10}}{1.879}+5 +\frac{1.5 \cdot 1.879}{2}=17.053m/s$

Next we write

$u_y=g \cdot t=9.81 \cdot 1.879=18.433m/s$

Ball speed before hitting a car

$u=\sqrt{u_x^2+u_y^2}=\sqrt{17.053^2+18.433^2}=25.111m/s$

The angle with the x axis is

$\alpha=\arccos{\frac{u_x}{u}}=\arccos{\frac{17.053}{25.111}}=47.226^0$