Find Time in Air for Ball Launched Upward with 18 m/s

AI Thread Summary
To determine the time a ball launched upward at 18 m/s is in the air, the Kinematic Equations of Motion should be used, specifically those applicable for constant acceleration due to gravity. The relevant equation relates initial velocity, acceleration, and time to calculate the total time of flight. Since air resistance is negligible, the only force acting on the ball is gravity. The ball will ascend until its velocity reaches zero before descending back to the ground. This approach provides a clear method for solving the problem.
iac0b
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A ball is launched directly upward from ground level with an initial speed of 18 m/s. (Air resistance is negligible.)

What equation would I use to find the time the ball was in the air?
 
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iac0b said:
A ball is launched directly upward from ground level with an initial speed of 18 m/s. (Air resistance is negligible.)

What equation would I use to find the time the ball was in the air?

Welcome to the PF.

Thread moved to Homework Help / Intro Physics.

You would use the Kinematic Equations of Motion for constant acceleration (since the acceleration due to gravity is constant on the Earth's surface). See half-way down this wikipedia page:

http://en.wikipedia.org/wiki/Kinematic_equations#Equations_of_uniformly_accelerated_motion

.
 
Thanks :)
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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