How to Calculate Projectile Motion with Initial Horizontal Velocity

AI Thread Summary
To calculate the time it takes for an object to hit the ground when pushed horizontally off a cliff, the vertical motion must be analyzed separately from the horizontal motion. The correct formula for vertical displacement is s = ut + 0.5at^2, where the initial vertical velocity (u) is 0, leading to a time of 3.0 seconds to hit the ground from a height of 45 meters. The horizontal distance traveled during this time, given an initial horizontal velocity of 2.0 m/s, is calculated to be 6.0 meters. The confusion arose from mixing horizontal speed with vertical acceleration, which was clarified in the discussion. Understanding the separation of horizontal and vertical components is crucial for solving projectile motion problems correctly.
Kotune
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Homework Statement


A night somebody pushes a derelict car horizontally off a 45m high cliff top at 2.0 ms^-1.
(a) How long does it take to hit the ground? (g=10ms^-2)
(b) How far from the base of the cliff does it land?


Homework Equations


The answer to (a) is 3.0s and (b) is 6.0m


The Attempt at a Solution


For (a) I tried s = ut +.5at^2. Hence 45 = 2t+.5(10)t^2 but the answer was not correct.
 
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Kotune said:
For (a) I tried s = ut +.5at^2. Hence 45 = 2t+.5(10)t^2 but the answer was not correct.

That's because you mixed a horizontal speed with a vertical acceleration.
 
Nugatory said:
That's because you mixed a horizontal speed with a vertical acceleration.

Ah ok I got it now, 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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