- #1
crism7
- 8
- 1
- Homework Statement
- j
- Relevant Equations
- t = 2vi sin theta / g (i'm not sure)
Do i need to find the x and y components??
j
Last edited:
Do i need to find the x and y components?
- Thread starter crism7
- Start date
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When a ball is shot at an angle of 55 degrees, its time in the air will be shorter compared to a vertical shot. The initial vertical velocity can be calculated by reducing the vertical component based on the launch angle. An algebraic expression for the initial vertical velocity is sufficient for the problem, rather than a numerical value. The given time of 1.34 seconds can be incorporated into the calculations to determine the impact of the angle on flight duration. Understanding these components is crucial for solving the problem effectively.
- #2
berkeman
Admin
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Welcome to PF. 
You are given how long the ball stays in the air when shot vertically. What is different when it is shot up at an angle of 55 degrees? It will stay in the air a shorter time, but by how much? What is the initial vertical velocity reduced to when the launch angle is 55 degrees?
You are given how long the ball stays in the air when shot vertically. What is different when it is shot up at an angle of 55 degrees? It will stay in the air a shorter time, but by how much? What is the initial vertical velocity reduced to when the launch angle is 55 degrees?
- #3
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My attempt:
Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE.
PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##.
PE of part in the tube = ##\frac{m}{l}(l - h)gh##.
Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##.
Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...