Velocity under constant acceleration

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nbroyle1
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The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45°. With what speed must the animal leave the ground to reach that height?

knowns: y=3.7 yo=0 Vo=0 V=want!

I tried finding the velocity by using the equation V^2=0+2(-9.8)(-3.7) but ended up with the wrong answer. What am I doing wrong?
 
on Phys.org
Do I need to get velocitys for the x and y components of the jump?? how do I do that?
 
This is projectile motion, so you have the break the velocity into components.
 
but I am not given a velocity so how do I break it into components?
 
The initial velocity is what you are looking. It's not zero.
the equation you want to use is ymax=((Vo^2)sin^2(theta))/2g
i believe
 
the solution i got is 10m/s if anyone else can verify
 
AmritpalS said:
the solution i got is 10m/s if anyone else can verify
Hmm, I'm obviously having finger trouble because I get nearer 12 m/s.
 
How do you know what the Vo is though?
 
yes the answer was 12m/s
 
The equation in the first post gives only the vertical component of velocity, you want the initial velocity.
 
sorry about that i plugged it into my calculator on radian mode. Should have just done it in my head. It is 12m/s sorry about that i guys
 
NemoReally said:
Hmm, I'm obviously having finger trouble because I get nearer 12 m/s.
no finger trouble is on me
 
nbroyle1 said:
How do you know what the Vo is though?
What you calculated in your original message was the vertical component of the puma's velocity.

The problem states that the puma's jumped at 45 deg from the horizontal and, as AmritpalS stated, you then need to work out what this works out as given the vertical component ... look at AmritpalS's equation and re-arrange it.
 
Ok thanks yea I was just trying to solve for the wrong variable. That equation is a little new to me Ill have to remember that one.