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A shot put is released with a velocity of 12m/s and stays in the air for 2.0s.
a)at what angle with the horizontal was it released?
b)What horizontal distance did it travel
A ball is hit at an angle of 45 degrees and goes 98m. Find the velocity of the ball when it left the bat.
I can't figure out a formula to give me these answers. Help is very appreciated!:)
Lets look at part a. Were looking for the angle that gives the ball a flight time of 2 seconds. The only factor in flight time for projectile motion is the y component of your velocity.Quote:
Originally Posted by 9665
Therefore
We also know from projectile motion that
where Vo is initial velocity, and Vf is final velocity. We know time, we also know Vfy, and we can substitute in for Voy and solve for the unknown angle!
If we know the angle at which this ball is traveling, we can determine the horizontal velocity of the ball by doing
and you can figure out horizontal displacement using
As for your other question, you need to use your kinetics formulas to figure out how long the ball needs to be in the air for. (Note its at a 45 deg angle, so your initial velocity in the y direction should be equal to the velocity in the x direction!)
If you need more assistance, don't be afraid to ask questions.:o
What's a kinetics formula?
Where did you get the second formula from?
What's the difference between Yo and Voy?
one more thing... so you mean that for every time there'e a 45degree angle Vx=Vy?
The kinetic formulas should be in your book, they allow you to represent motion as a function of time.Quote:
Originally Posted by 9665
The second formula is one of the kinetic equations available.
Yo is the initial vertical position, Voy is the initial vertical velocity.
Yes, in any 45 degree corner, the adjacent and opposite components to that angle are equal in magnitude.
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