Solving Mouse Dropped by Owl - 6.3m Altitude

  • Thread starter Thread starter omc1
  • Start date Start date
  • Tags Tags
    Bird
AI Thread Summary
The discussion focuses on calculating the time it takes for a mouse to hit the ground after escaping from an owl at an altitude of 6.3 meters. The owl is flying at a speed of 6.5 m/s at an angle of 11 degrees above the horizontal. The initial vertical velocity component is derived using the sine function, which is crucial for accurate calculations. The user initially struggled with the quadratic equation but realized the importance of considering only the vertical component of velocity. The conversation highlights the significance of understanding vector components in projectile motion problems.
omc1
Messages
100
Reaction score
0

Homework Statement

Having captured a mouse, an owl is flying towards the barn where it lives to enjoy dinner. While approaching the barn at a speed of 6.5 m/s at an upwards angle of 11o above the horizontal, the struggling mouse escapes from the talons of the owl while at an altitude of 6.3 m. How much times passes before the mouse hits the ground?

Homework Equations

y=ynot+ vnot Δt +1/2aΔt



The Attempt at a Solution

i used my vnot 6.5m/s and v 6.5sin11. then did the quadratic equation to find Δt but i keep getting the wrong answer.
 
Physics news on Phys.org
The velocity is at an angle. The time to the ground depends on its vertical component, not its entire magnitude.
 
oh ok got it now thanks!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top