Physics Help Forum Projectile Motion - Hitting a Sloped Wall

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Mar 29th 2018, 10:24 PM #1 Junior Member   Join Date: Mar 2018 Posts: 2 Projectile Motion - Hitting a Sloped Wall Question: In the figure, a ball is launched with a velocity of magnitude 7.00 m/s, at an angle of 43.0° to the horizontal. The launch point is at the base of a ramp of horizontal length d1 = 6.00 m and height d2 = 3.60 m. A plateau is located at the top of the ramp. (a) Does the ball land on the ramp or the plateau? When it lands, what are the (b) magnitude and (c) angle of its displacement from the launch point? The ball does in fact land on the ramp; not making it to the plateau. However, I don't know how to find the x and y distances of the impact. I have a feeling that I need to use the [x = Velocity * cos(theta)*time] equation and plug it into my [y=y0 + ut + 1/2gt^2] equation to solve for time, but i'm just not getting the right answer. Any suggestions?
 Mar 30th 2018, 05:13 AM #2 Senior Member     Join Date: Jun 2016 Location: England Posts: 1,003 Your equation [y=y0 + ut + 1/2gt^2] is correct but... Note that [u] in the above equation is the initial vertical component of the velocity and (y0 is zero). let us call the vertical component of velocity uy so [uy=7.0*sin(43.0°)] Also the ramp is rising according to the equation [y=d2/d1*x] where x is the horizontal distance moved by the ball. x will depend on the horizontal velocity of the ball let us call it ux so [ux=7.0*cos(43.0°)] Now we can see that: [x=ux*t] this gives us 2 equations for y: [y=d2/d1*7.0*cos(43.0°)*t] and: [y=7.0*sin(43.0°)t + 1/2gt^2] At what t (other than zero) do these two equations give the same answer? What is x at this time? is this less than 6m? topsquark likes this. __________________ ~\o/~ Last edited by Woody; Mar 30th 2018 at 05:20 AM.

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