Intial speed of projectile motion from height

AI Thread Summary
To calculate the initial velocity of a projectile dropped from a height without using time, one can apply kinematic equations that separate horizontal and vertical motion. The horizontal motion maintains a constant velocity, while vertical motion is influenced by gravitational acceleration. For a projectile dropped from a height of 2.8 meters that travels a horizontal distance of 3 meters, the relevant equations involve initial vertical velocity set to zero and the final vertical velocity calculated using gravitational acceleration. The key is to relate the vertical displacement to the horizontal distance using these principles. Understanding these concepts allows for determining the initial velocity without directly calculating time.
ghostbuster25
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ok I am trying to find out how to work out the intial velocity of a projectile dropped from a height h and reaches a distance s.

say a ball thrown from a height of 2.8 metres and reaches a distance of 3 metres.

I want to be able to work it out WITHOUT having to work out the time.
so can't use s=v*t

does anyone kow What is the equation for this?
 
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ghostbuster25 said:
ok I am trying to find out how to work out the intial velocity of a projectile dropped from a height h and reaches a distance s.

say a ball thrown from a height of 2.8 metres and reaches a distance of 3 metres.

I want to be able to work it out WITHOUT having to work out the time.
so can't use s=v*t

does anyone kow What is the equation for this?

You must know these concept. All the vectors you must consider 2 components. Horizontal Motion velocity is constant and vertical motion acceleration is constant.If the particle dropped from the hill, no need to consider angle for initial velocity.
Horizontal Motion Vertical motion
initial velocity v(0x)= v(0) v(0y)=0 (for above case)

final velocity v(x)=v(0x)=v(0) v(y)=...

acceleration a(x)=0 a(y)=-g

displacement x y

time t t (scalar)

x=v(0x)t y=v(0y)t -0.5 gt^2
v(y)=v(0y) -gt
v^2=v(0y)^2-2gy
You may use simple free fall bodies equations.
 
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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