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An object is launched from a semicircular chute 1.4 meters from the ground at a constant speed. The radial acceleration is 7.9 m/s^2 and radius is 0.7 m. Find the horizontal range the object travels from the bottom of the chute to the moment it hits the ground.

My attempt: Given that $a_r=7.9 \ m/s^2$ then $v_0=2.35$. I know that $a_y=-g, a_x=0$ and the ball hits the ground $t_f=0.825$ seconds after the launch.

So the positive along the y direction is given by $y(t)=1.4+2.35t-4.905t^2$, but it’s position along the x direction is $x(t)=v_{0x}t$ where $v_{0x}=v_x$. Thus the range is $0.825v_{0x}$

But I cannot find out what $v_{0x}$ is. The most I know at the moment is that $2.35=v_{0x}+v_{0y}$

Lex_i
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1 Answers1

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A semicircular chute implies all the velocity is actually in the horizontal direction. So the time your object hits the ground is $t = \sqrt{2y/g} = \sqrt{1.4/4.9} = 0.534$ seconds, and not $0.825$. The range is then just $v_{ox}t = v_ot = 0.534*2.35 = 1.25m$

cybershiptrooper
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