Answer in Mechanics | Relativity for lili #108279
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”
b) 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”
