Projectile Motion: Salmon Jumping a Waterfall - Calculating Minimum Speed Needed

[Rumplestiltskin]

Hi all. I just recently dropped psychology for physics so this topic as a whole has me stumped and the textbook isn't helping. I'll be using this thread for any questions that arise pertaining to projectile motion (so more than one), if that's cool? Should be really basic. Thanks.

1. Homework Statement
A salmon moving upstream to its breeding grounds jumps a waterfall 2.5m high. With what minimum speed must it leave the water below to reach the top level?

Homework Equations

[/B]
Vectors:
V_y = Vsinθ
V_x = Vcosθ
Suvat?

3. The Attempt at a Solution
Don't understand how you can work anything out with one piece of data...

 

[gneill]

Hi Rumplestiltskin, Welcome to Physics Forums.

I'll be using this thread for any questions that arise pertaining to projectile motion (so more than one), if that's cool? Should be really basic. Thanks.

That would be contrary to forum rules. One problem per thread, so new questions get new threads (even if they are on related topics).

Homework Statement
A salmon moving upstream to its breeding grounds jumps a waterfall 2.5m high. With what minimum speed must it leave the water below to reach the top level?

Homework Equations

[/B]
Vectors:
V_y = Vsinθ
V_x = Vcosθ

3. The Attempt at a Solution
Don't understand how you can work anything out with one piece of data

You're given more than you might think as there is the assumption that this is a projectile motion problem taking place near the surface of the Earth. So you know, for example, that the motion will be affected by the acceleration due to gravity (g). and that all the standard kinematic equations for projectile motion apply.

Some things to become familiar with in the study of projectile motion include the range formula, the launch angle for maximum range, the maximum height of a projectile given its launch conditions (speed, angle).

 

[Rumplestiltskin]

After a cursory reading into the topics you mentioned I came across the vaguely familiar formula v = √2gs. Plugging in 9.81ms^-2 for g and 2.5m for displacement s, I arrived at 7 ms^-1 for the answer; the mark scheme confirmed this. Thanks! And sorry to waste your time.