Solving Velocity Problem: 26°, 23m, 12m, 4.8s

[Lordrunt]

Homework Statement

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?

Homework Equations

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

The Attempt at a Solution

I wish I could, but I've only used 45° problems like this.

 

[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 V_x and V_y be?
Regarding the equations you quote:

X=V_x + t​

That should be X=V_x t

Y=V_yt + 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.

 

[Lordrunt]

26 degrees upward from horizontal

 

[haruspex]

26 degrees upward from horizontal

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

 

[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