Downward, find airtime, velocity

  • Thread starter Thread starter bigman8424
  • Start date Start date
  • Tags Tags
    Velocity
AI Thread Summary
The discussion centers on calculating the airtime and impact velocity of a water balloon thrown downward at 15 m/s from a height of 25 m. Participants emphasize the importance of understanding the physics concepts before applying equations, specifically questioning the meaning of "air time" and what factors determine its end. There is confusion regarding the use of the equation, particularly the value of gravitational acceleration and the relevance of the sine function in this context. The correct approach involves solving for the time when the balloon reaches the ground, rather than misapplying the kinematic equation. Accurate calculations require clarity on the direction of the initial velocity and the correct value for gravitational acceleration.
bigman8424
Messages
25
Reaction score
0
water balloon thrown down at 15m/s to ground 25 m below, what is air time and velocity at impact
y = y(i) + v(i)sint + 1/2gt2
y = 25 + 15sin(90)t + 1/2(-9.8)t2 ? ?
 
Physics news on Phys.org
dude, try to understand the physics before pluging in numbers into a meaningless equation...read the book first...
 
1. What does "air time" MEAN?
2. More specifically, what ends the "air time"?
3. What is y when the "air time" ends?
4. No, not "y = y(i) + v(i)sint + 1/2gt2 "
5. Do you know what sin(90) is?

6. You need to solve an equation to find the time t when "air time" ends.
 
first of all, g is not -9.8. you throw it downwards, not upwards
 
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}...
Back
Top