Kinematic equation problem -- A fountain shoots water out of the ground ....

AI Thread Summary
A fountain shoots water at a 40° angle with a horizontal velocity of 2 m/s, landing 3 meters away. The user attempts to calculate the maximum height using two kinematic equations but receives conflicting results. The first method yields a negative height, indicating an error, while the second method appears to provide a more plausible answer. A key insight reveals that a velocity of only 2 m/s cannot achieve a 3-meter horizontal distance, suggesting the problem's conditions are unrealistic. Ultimately, the second method is deemed correct, as it does not rely on the impossible range information.
dragon-kazooie
Messages
16
Reaction score
0

Homework Statement


1. A fountain shoots water out of the ground at a 40° angle. It’s horizontal velocity is 2 m/s and the water lands 3 meters away. What is the maximum height that the water reaches during that time?

Homework Equations


Dy =vit + ½ at2

and

Vfy2 = Viy2 + 2ady

The Attempt at a Solution



First solve for the time in the air: t = dx /vx t = 3m/2m/s = 1.5s
Divide that in half to find the time at maximum height: t = 0.75s

Solve for initial vertical velocity:
Tan θ = Vy /Vx
Tan 40° = Vy /2m/s
Vy = 1.68 m/s

Now I get two different answers when I use different equations, and I don't see why. Using this one that I did in purple must be wrong, because it doesn't make sense for ½ at2 to be a larger value than vit and I get a negative number which can't be right. But I don't see where the mistake is! What am I doing wrong?

dy =vit + ½ at2
dy =1.68 × 0.75s + ½ -9.8m/s2 × (0.75s)2
dy = 1.26 - 2.75
dy = -1.49m

If I use this formula in blue I get a different final answer which seems more correct:
Vfy2 = Viy2 + 2ady
0 = 1.6782 + 2 × -9.8 × dy
0 = 2.816 + 2 × -9.8 × dy
-2.816 = -19.6 dy
0.143 = dy

Is the one in blue correct? What did I do wrong with the one in purple?

Thank you in advance!
 
Last edited by a moderator:
Physics news on Phys.org
@dragon-kazooie , your first method is correct. You got a crazy answer because the given conditions are impossible. There is no way that a jet of only 2m/s can reach 3m horizontally, even at 45 degrees.
Your second method appeared to work because you do not need the 3m range information, and you effectively ignored it.
 
Last edited by a moderator:
  • Like
Likes dragon-kazooie
Thank you @haruspex! That was driving me crazy!
 
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}...
View full post »

Similar threads

How Do You Calculate the Initial Velocity of Water from a Drinking Fountain?
Replies
1
Views
6K
Free Fall and a Projectile motion question
Replies
10
Views
2K
Work problem -- lifting water out of tanks
Replies
2
Views
2K
Difficult Projectile Motion Arrow Problem
Replies
7
Views
4K
Six kinematics problems - all either worked out with wrong answer or conceptual
Replies
3
Views
12K

Hot Threads

Recent Insights

Back
Top