VERY DIFFICULT 2-D Motion Problems

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The discussion focuses on solving two complex 2-D motion problems involving projectile motion. For the first problem, participants suggest calculating the vertical component of the ball's velocity at the moment it was caught to determine its maximum height and initial velocity. In the second problem, the horizontal distance traveled and the maximum height reached are used to find the initial velocity components of the basketball. Participants emphasize breaking down the problems into horizontal and vertical components and using kinematic equations to solve for unknowns. The conversation highlights the importance of organizing known values and using a systematic approach to tackle these physics problems effectively.
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Homework Statement



1. A baseball player hits a homerun and the ball is caught by a person in the stands. It is caught 7.50m above the point it was hit. At the moment it was caught, it has a velocity of 36.0m/s at an angle of 28 degrees below the horizontal. What was the initial velocity of the ball when it was hit.

2. Basketabll hoop is 3.05 m above the playing surface. A basket is made. The ball reached a max. height that was 2.00m above the height of the basket hoop. The basketball was launched from a height of 1.95m. If the ball traveled a horizontal distance of 5.20 m in 2.00 seconds, what was the initial velocity of the basketball?



Homework Equations


v f = vi + at
delta x= vi + 1/2 a T^2
vf ^2= Vi^2+ 2 a delta x


The Attempt at a Solution


I have no idea where to start. Can someone lead me down the right path to answering these questions?
 
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For number 1, you can calculate the vertical component of the ball's velocity at the moment it was caught, and from that can easily deduce from what maximum height it began to descend. The maximum height reached tells you the initial vertical component of the velocity (upon being hit).
 
For number 2, the time taken to travel a horizontal range of 5.20 m tells you the initial horizontal component of the velocity. The max height reached tells you the vertical component of the initial velocity.
 
I don't understand what you mean by that. Can you please explain it a bit further?
 
In problem 1, you know the magnitude and direction of the velocity vector of the ball at the moment it is caught. Therefore, you can resolve the velocity into horizontal and vertical components. The vertical component (how fast it was traveling "downwards") tells you what maximum height it fell from (measured above the position at which it was caught), because you know that it was accelerating under gravity.
 
Vyo=Vosin(theta)
Vxo=Vocos(theta)

Right down all your knows in a t chart. one side you x's and the other your y's

ay=-9.8
delta y=7.50m
Vo=36m/s
There are some more
x side equation is(Delta)x=VoT
Y equation (Delta)y=Vyot+1/2Ayt^2

Solve for time then you can plug that in and get Vyo
 
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