# Homework Help: Quadratic Projectile

1. Feb 4, 2012

### PonderingMick

1. The problem statement, all variables and given/known data

Motorbike crossing a river using a ramp. River is 40m wide, the bank on the opposite side is 15m lower. The ramp angle is 53 degress. What is the speed needed at take off?

2. Relevant equations
I am using s = ut for the horizontal
and s = ut x 1/2 at2
I think because the bank is lower the other side I need to use the quadratic equation?

3. The attempt at a solution
Using s = ut I get ut = 50/cos 53
Which I then substitue in s = ut x 1/2 at2
Which gives:
15 = 40/cos53 + 1/2 at2
Which i rearrange:
4.92 + 40/cos53 -15 = 0
I then try and use the quadratic equation:
t = (-66√662-4 * 4.9 * -15) /9.8
Which gives 0.22s or -13s

Am I anywhere near the solution?

I know the answer is 19.8 m/s

Last edited: Feb 4, 2012
2. Feb 4, 2012

### PhysicsGente

Are you sure the answer is 19.8 m/s?

3. Feb 4, 2012

### PonderingMick

Yes, the answer is in my textbook

4. Feb 4, 2012

### AlchemistK

Check your directions.Gravity acts downwards, you've substituted g without a negative sign.

5. Feb 4, 2012

### PonderingMick

t = (-66√662-4 * 4.9 * -15) /9.8
So I use -4.9 instead
t = -66 + or - ( √662 -4 * -4.9 * -15) /9.8
t = -66 + or - ( √4356 - 294) /9.8
t = -66 + or - (66-294) /9.8
t = -66 + or - -23
t = -66+-23 = -89 or -43

t must be +ve?

6. Feb 4, 2012

### AlchemistK

The bank on the other side is 15 m LOWER. So the displacement is actually "-15" m.

You don't have to choose the conventional directions like g acts downwards, but whatever you choose, make sure you do choose something, and follow it throughout the question. If "up" is positive, it should be positive throughout.

7. Feb 5, 2012

### PonderingMick

Just checked the answer and its actually 17.8 ms-1, will try and work through it again later.