Question #108236

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

question is in the picture, also d8=4

question is in the picture, also d8=4

Expert's answer

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"

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