# Projectile Motion Problem with spitting a seed

1. Jan 4, 2009

### cmh_580

1. The problem statement, all variables and given/known data
The record for spitting a watermelon seed by a man is 68.75 feet. Assuming he spat at an angle of 45.0 degrees and from an initial height of 1.50 meters above the ground, how fast did he spit the seed out of his mouth. (neglect any air resistance)

2. Relevant equations
The only equations I know are basic kinematic equations. No mass/force stuff. Some of the equations we use are:
d-d0 = V0t + 1/2at^2
V = V0 + at
V^2 = V0^2 + 2a(d-d0)

That's pretty much it. The 0's should be subscripts, but I don't know how to do that.

3. The attempt at a solution
I know how to solve for the time it takes to reach its max height (t=(V0sin45)/(a)), but from there I am stuck.
For the time it takes to go down from its max height, I know what V0 is and I know what a is, but that's it, so I don't know how to solve for that. I think if I'm going to get initial velocity I first need 3 other variables. And unless I get time, I can't have 3, right?

I don't just want the answer, because that doesn't really help me at all. A simple explanation (I've only been in Physics for less than 4 months) would be very helpful.

Last edited: Jan 4, 2009
2. Jan 4, 2009

### Hootenanny

Staff Emeritus
Welcome to Physics Forums.

A good place to start would be to calculate the total flight time, that is the time taken for the seed to reach it's maximum height and fall back to earth. Can you do that? Note that the initial and final heights are not the same.

Hint: The vertical and horizontal components of the motion are independent, which means you can consider the vertical motion of the speed whilst ignoring the horizontal motion (or vis versa).

3. Jan 4, 2009

### cmh_580

I don't know how to solve for the time for it to fall back to earth. If I only have two variables then I'm not sure how to do it. In order to solve for time I would also need to know either the final velocity or the vertical displacement, and I have neither of those. The vertical displacement would be the distance to the top plus 1.5 meters, but I don't know what the distance to the top is.

4. Jan 4, 2009

### Hootenanny

Staff Emeritus
But you do have the vertical displacement! What is the seed's initial [vertical] position (d0)? What is it's final [vertical] position (d)?

5. Jan 4, 2009

### cmh_580

So the vertical displacement would be only -1.5m? But that doesn't account for when it went up in the air and came back down does it?

6. Jan 4, 2009

### Hootenanny

Staff Emeritus
Correct! No, it doesn't account for the fact that it went upwards first, but the initial velocity does

On a side note you will obtain two flight times, only one will correspond to the problem in hand. The other will be s spurious solution corresponding to the section of the parabola for negative time.

7. Jan 4, 2009

### cmh_580

Okay, so here's what I did. Assuming the following:
vertical acceleration = -9.81 m/s^2
initial vertical velocity = V0sin45 m/s
vertical displacement = -1.5 m

horizontal acceleration = 0 m/s^2
initial horizontal velocity = V0cos45 m/s
horizontal displacement = 68.75 m

Okay, so using the horizontal numbers I can put t in terms of V0:
d-d0 = V0t (a=0, so that cancels out the other part)
t=(d-d0)/(V0)
t=(68.75)/(V0cos45)

So that's t, which is the same for both vertical and horizontal

From there, I put that t into the equation:
d-d0 = V0t + 1/2at^2
using the vertical numbers for those variables, I came up with this
-1.5 = (V0sin45)*(68.75/V0cos45) + 1/2(-9.81)((68.75/V0cos45)^2)

After multiplying/dividing out the numbers that I could, I was left with this:
(V0cos45)^2 = 330.0183496
Finding the square root of both numbers, I got
V0cos45 = 18.16641
Then, after dividing by cos45 (.707), I got
V0 = 25.69179 m/s, which would reduce to 25.7 based on sig figs.

Please tell me this is right, if not, where did I go wrong?

8. Jan 4, 2009

### Hootenanny

Staff Emeritus
I have V0 = 26.0 m/s to 3sf.
You're good up to here. I'm not sure whether you've made an algebraic slip, or have simply rounded too early.

9. Jan 4, 2009

### cmh_580

I did it again and still got 25.7. Either way, at least I know I'm doing it right. Thanks a lot for the help.