# Homework Help: Solving for velocity

1. Nov 29, 2012

### Lordrunt

1. The problem statement, all variables and given/known data
A boy jumped off of a cliff into the water at a 26° angle. He traveled 23 meters and fell 12. He had 4.8 seconds of travel time. What was the velocity needed to do this?

2. Relevant equations
X=Vx + t
Y=Vyt + 1/2at^2

3. The attempt at a solution
I wish I could, but I've only used 45° problems like this.

2. Nov 29, 2012

### haruspex

Is the angle measured from vertically up, upwards from horizontal, downwards from horizontal, or from vertically downwards? Assuming it's θ above horizontal, if the take-off speed is V, what would Vx and Vy be?
Regarding the equations you quote:
X=Vx + t
That should be X=Vx t
Y=Vyt + at2/2
Need to be careful with the signs. First, define whether up or down is your positive direction, then use that consistently for distance, speed and acceleration.

3. Nov 29, 2012

### Lordrunt

26 degrees upward from horizontal

4. Nov 29, 2012

### haruspex

So what would Vx and Vy be? (Preferably expressed in terms of an arbitrary angle θ, rather than specifically 26o.)

5. Nov 30, 2012

### johns123

distance = Vx . t where t = 4.8sec and distance in x-direction = 23 meters

so Vx = 23 meters / 4.8sec = 4.79 m/s

finally V = Vx / sin23 = 12.26 m/s assuming angle measured from verticle