Finding Energy while knowing intial Y and desired x

  • Thread starter Thread starter sfishman59
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
To determine the energy required for an object launched from a height of 1.7526 meters to reach a horizontal distance of 1.8288 meters, the initial speed and angle of launch must be defined. The hypotenuse calculated as 2.532888 meters is not relevant to the problem. Instead, the motion should be analyzed using the equations of constant acceleration in both the horizontal and vertical directions. The discussion emphasizes the importance of separating the x and y components of motion to solve for the initial speed and angle. Accurate calculations will lead to the required energy exerted for the launch.
sfishman59
Messages
1
Reaction score
0

Homework Statement



An object is launched at x m/s from a 1.7526 meter platform. How much energy must be exerted for the object to be launched 1.8288 meters?

Homework Equations



Not Sure

The Attempt at a Solution



I made a right triangle and found that the hypotenuse (or distance that the object will need to travel) is 2.532888 meters.
 
Physics news on Phys.org
welcome to pf!

hi sfishman59! welcome to pf! :wink:

do you mean that the object is launched at an unknown angle from 1.7526 m above the ground, and has to land on the ground 1.8288 m away?

the hypotenuse has nothing to do with it

call the initial speed v and the initial angle θ, and use the standard constant acceleration equations in the x and y directions (separately, with a = 0 and a = -g) …

show us what you get :smile:
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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}...
Back
Top