A person on a small, 7.00 meter high, hill aims a water-balloon slingshot straight at a friend hanging from a tree branch 4.00 meters above and 6.00 meters away. At the instant the water balloon is released at 7.00 m/s, the friend lets go and falls from the tree, hoping to avoid being hit by the water balloon. Calculate the following:
(a) the time for balloon to reach tree ______________s
(b) the height of balloon at that time ________________m
(c) the height of the falling friend at that time ___________________m
i don't even know where to begin. thanks!
Draw a picture.
It is not clear to me where the 7m hill comes in. If I interpreted it correctly, it is of no use except that it permits the water balloon and the person to fall freely below the ground!
See:
http://img689.imageshack.us/img689/5825/1260658746b.png
Let
D=horizontal distance between the two friends = 6m
V=vertical distance between the two friends = 4m
u= initial velocity of the balloon making an angle θ (theta) with the horizontal = 7 m/s
ux=horizontal component of initial velocity = ucos(θ)
uy=vertical component of initial velocity = usin(θ)
H1(t)=height above initial level of balloon at time t
= uy*t-(1/2)gt²
H2(t)=height of friend measured from top of tree at time t
= 0 -(1/2)gt²
1. Time to reach tree
T = D/ux
2. H1(T)=uy*T-(1/2)gT²
3. H2(T)=-(1/2)gT²
Note that H1 is measured from ground, and H2 is measured from top of tree, at 4 m above ground.
My calculations tell me that the balloon hits the poor friend.
Forgot to mention:
θ=(theta) is the angle aimed directly at the friend, so
θ=arcTan(4m/6m)
To solve this problem, we can start by breaking it down into several steps.
Step 1: Calculate the time for the balloon to reach the tree branch:
We can use the equation of motion for vertical motion to find the time it takes for the balloon to reach the tree. The equation is:
h = vi * t + (1/2) * g * t^2
where:
- h is the height (7.00 meters)
- vi is the initial vertical velocity (0 m/s when it is released)
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time
Rearranging this equation and substituting the given values, we get:
0 = -4.9t^2 + 7t
This equation is a quadratic equation. We can solve it by factoring or using the quadratic formula.
Step 2: Calculate the height of the balloon at that time:
To calculate the height of the balloon at the time it reaches the tree branch, we can use the equation of motion again:
h = vi * t + (1/2) * g * t^2
In this case, the initial vertical velocity is 7.00 m/s (as the balloon is released at that speed) and time is the value we found in step 1.
Step 3: Calculate the height of the falling friend at that time:
To calculate the height of the falling friend, we need to find the time it takes for the friend to fall from the tree to the ground. We can use the equation:
h = vi * t + (1/2) * g * t^2
In this case, the initial vertical velocity is 0 m/s (as the friend lets go from rest), and the height is 4.00 meters (the height of the tree branch).
Again, we can rearrange and solve the equation to find the time.
I will calculate the values for you.