Physics Question Help Please

[moto_za]

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A bottle is thrown with an initial velocity of 4m/s at 45 degree from the horizon. Find its final horizontal and vertical velocities before striking the ocean. TID. 🙂

 

[loveoforganic]

Sounds like a hw prob. Break the velocity into vertical and horizontal components, you have your answers. Horizontal velocity is the same throughout, and total energy isn't going to change without height change, so your vertical component will be the same at t zero and tfinal

 

[Blaughable]

If it's being thrown at the same height as to where it's landing then the speed and angle going up will be the same as the speed and angle going down at any horizontally drawn line without any air resistance. So, you would just find the x and y components of 4m/s at -45degs ( 2.82m/s in the +x direction, and 2.82m/s in the -y direction)

You should know that at 45deg the components are hypotenuse*.707 or know that the components are hypotenuse*sqrt(2)/2 and know that the sqrt(2) is 1.41 and do math from that.

My method would be:

components = hypotenuse*sqrt(2)/2

=4*sqrt(2)/2 = 2*sqrt(2) = 2*1.41 = 2.82

 

[blue0rchid]

very simple.
without air resistance, initial velocity = final velocity for any projectile motion (given that they start and end on the same plane). So final velocity is 4m/s at 45 degree downward (south east). sin(45)=cos(45)=sqrt(2)/2. So vx = vy = v0 * sin(45) = 4 * sqrt(2) / 2 = 2 sqrt(2)