Calculate Speed for Motorbike Crossing a 40m Wide River Using a 53 Degree Ramp

[PonderingMick]

Homework Statement

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?

Homework 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?

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

 

[PhysicsGente]

Are you sure the answer is 19.8 m/s?

 

[PonderingMick]

Yes, the answer is in my textbook

 

[AlchemistK]

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

 

[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?

 

[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.

 

[PonderingMick]

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