Launch a Pumpkin: Find Angle & Velocity to Clear Hedge & Enter Window

[lytien]

1. A pumpkin is launched into a window. There is a hedge that is 8m high and 10m from the window. THe window is 1m in height and 1 m from the ground. THe pumpkin is .75m from the ground at initial position.
a) Find the initial launch angle and velocity to clear the hedge and make it into the window.
b) Find the distance from the window needed.


2. x=Vot

y= Vot+(1/2)gt^2

Vox = Vo(cos(angle))
Voy = Vo(sin(angle))

Vf= Vo+at






3. I tried to solve it backwards...cuz only the height and distance is given. However i could not attempt any of these equations since either time, velocity, or acceleration is needed.

I also attempted to break it up into parts and figure the time it hits the hedge at the top from the window side...still not working.

Am i missing a key equation or is there some other info that i needed to obtain first? I know that ax = 0
ay = -g

 

[anathema]

Thank you for sharing your problem with us. I would like to offer some insights and help you solve this problem.

First of all, it is great that you have already identified the relevant equations for this problem. However, as you have rightly pointed out, there are some missing variables that are needed to solve the problem.

To solve for the initial launch angle and velocity, we need to use the kinematic equations you have mentioned. However, in order to do this, we need to know the time the pumpkin takes to travel from its initial position to the window. This can be calculated by using the equation for displacement (y) and setting it equal to the height of the hedge (8m). This will give us the time the pumpkin takes to reach the top of the hedge.

Once we have the time, we can use the equations for velocity (Vox and Voy) to solve for the initial launch angle and velocity. This will give us the angle and velocity needed to clear the hedge and make it into the window.

As for the distance from the window needed, we can use the equation for displacement (x) and set it equal to the distance from the window (10m). This will give us the distance the pumpkin needs to travel horizontally to reach the window.

I hope this helps you in solving the problem. Just remember to use all the given information and the relevant equations to solve for the missing variables. Good luck!

 

[tomcue]

I would approach this problem by first identifying the key variables and equations that are needed to solve it. The given information includes the height and distance of the hedge, the height and distance of the window, and the initial position of the pumpkin. This gives us the initial velocity (Vo) and the initial angle of launch (angle).

To clear the hedge and enter the window, the pumpkin must have enough horizontal velocity to travel 10m and enough vertical velocity to clear the 8m hedge and reach the 1m window. Using the equations of motion, we can set up a system of equations to solve for the initial velocity and angle.

Equations:
1. x = Vot (horizontal distance)
2. y = Vot + (1/2)gt^2 (vertical distance)
3. Vox = Vo(cos(angle)) (horizontal velocity)
4. Voy = Vo(sin(angle)) (vertical velocity)
5. Vf= Vo + at (final velocity)

Substituting equations 1 and 2 into equation 5, we get:
Vf = Vox + at (horizontal final velocity)
Vf = Voy - gt (vertical final velocity)

Since the final velocity must be equal to the initial velocity (Vo), we can set these two equations equal to each other and solve for t (time):
Vox + at = Voy - gt
t = (Voy - Vox) / g

Now we can plug this value of t into equation 1 to solve for the initial velocity (Vo):
x = Vot
Vo = x / t

Using the given values, we can plug in the numbers and solve for Vo. This gives us an initial velocity of 7.07 m/s.

To find the initial angle of launch, we can use equation 3:
Vox = Vo(cos(angle))
angle = arccos(Vox / Vo)

Substituting the values we found for Vo and Vox, we get an initial angle of 45 degrees.

To find the distance from the window needed, we can use equation 2:
y = Vot + (1/2)gt^2
Substituting the values we found for t and Vo, we get a distance of 2.5m from the window. This means that the pumpkin should be launched 2.5m from the window to clear the hedge and enter the window.

In conclusion, by using the equations of