Do i need to find the x and y components?

AI Thread Summary
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.
crism7
Messages
8
Reaction score
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:
Physics news on Phys.org
Welcome to PF. :smile:

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?
 
To @crism7 : Note that @berkeman is not expecting you to find a number for the initial vertical velocity. An algebraic expression will be sufficient. Then see how you can work the given 1.34 s into it.
 
  • Like
  • Informative
Likes berkeman and crism7
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Back
Top