Calculating the Height and Time of a Kicked Stone

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To calculate the height of the bridge, use the horizontal velocity of the stone (3.5 m/s) and the horizontal distance it traveled (5.4 m) to determine the time of flight, which is approximately 1.54 seconds. During this time, the stone falls under gravity, allowing the height to be calculated using the formula for free fall. If the stone were kicked harder, increasing its horizontal velocity, the time to fall would remain unchanged, as the vertical motion is independent of horizontal motion. The key takeaway is that the height can be determined from the time of flight and gravitational acceleration. Understanding these principles is essential for solving projectile motion problems.
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Homework Statement



Bill stands on a bridge kicking stones into the water below.

A) If Bill kicks a stone with a horiz velocity of 3.5 m/s and it lands in the water a horiz distance of 5.4 m from where he is standing, what is the height of the bridge?

B) IF the stone had been kicked harder, how would this affect the time it would take to fall?


Homework Equations





The Attempt at a Solution



Don't know how to begin.
 
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From the horizontal velocity and horizontal distance, one can compute the time traveled.

During the horizontal travel (same time), the stone also falls under the influence of gravity. How far does it fall in that time?


A good reference -

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html


The time of falling is the temporal constraint.
 
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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