Horizontal/verticle distance, velocity

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To find the horizontal distance a water bomber's load travels before hitting the ground, the time of fall is calculated using the formula t=√(2d/g), where d is the altitude (300 m) and g is the acceleration due to gravity (approximately 9.81 m/s²). This results in a time of about 7.82 seconds. The horizontal distance is then calculated using d=vt, where v is the velocity of the bomber (60 m/s). The final distance should be converted from meters to kilometers by dividing by 1000, not 60. The mistake in the calculation was misunderstanding the conversion factor for meters to kilometers.
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Homework Statement


At an altitude of 300 m, a water bomber drops a load of fire repellent on a hot spot of a forest fire. If the bomber is flying at a velocity of 60 m/s, what horizontal distance (km) does the load travel before hitting the fire?

Homework Equations


d=v(t), t=√(2d/2)


The Attempt at a Solution


i tried finding time using the formula i gave above which gave me 7.82 seconds. then i tried plugging that and 60m/s into d=vt then diving that answer by 60. ( to put it into km). what am i doing wrong?
 
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Why divide by 60? If you are converting from meters to kilometers, remember that there are 1000 meters in a kilometer...
 
o goodness. what a stupid mistake. thank u so much
 
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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