- #1

I am having trouble figuring out this problem:

A ball is launched with a velocity of 10 m/s at an angle of 50 deg 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 b) the magnitude and c) the angle of its displacement from the launch point?

The diagram that I was given looked like this

--------/|--------

-------/ |

------/--d2

-----/---|

___o/-d1--

note: please ignore the ---- along the left side of the ramp, it was the only way i could figure out how to create the picture. the o = the ball. it's on the ground. the --- lines to the right of the ramp is the plateau.

I was using the trajectory equation:

y = (tan@)x - gx^2/2(Vocos@)^2

= (tan50)(6) - 9.8(6)^2/2(10cos50)^2

= 2.88 m

Then I found the length of the ramp:

ramp = d = (3.6^2 + 6^2)^(1/2) = 36.5 m

With this in mind, I was thinking that the ball would land on the ramp because 2.88 m does not even come close to clearing the length of the ramp, 36.5 m. So, would the magnitude of the displacement be 2.88 m? Concerning the angle, why would it be any different from 50 deg? I think that I'm getting confused about what they're looking for. Could someone help me understand this, please? Thanks so much

A ball is launched with a velocity of 10 m/s at an angle of 50 deg 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 b) the magnitude and c) the angle of its displacement from the launch point?

The diagram that I was given looked like this

--------/|--------

-------/ |

------/--d2

-----/---|

___o/-d1--

note: please ignore the ---- along the left side of the ramp, it was the only way i could figure out how to create the picture. the o = the ball. it's on the ground. the --- lines to the right of the ramp is the plateau.

I was using the trajectory equation:

y = (tan@)x - gx^2/2(Vocos@)^2

= (tan50)(6) - 9.8(6)^2/2(10cos50)^2

= 2.88 m

Then I found the length of the ramp:

ramp = d = (3.6^2 + 6^2)^(1/2) = 36.5 m

With this in mind, I was thinking that the ball would land on the ramp because 2.88 m does not even come close to clearing the length of the ramp, 36.5 m. So, would the magnitude of the displacement be 2.88 m? Concerning the angle, why would it be any different from 50 deg? I think that I'm getting confused about what they're looking for. Could someone help me understand this, please? Thanks so much

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