Finally! Now I understand why the acceleration is zero in the x direction! And even, I should have put it together on my own, now I understand why that means the x direction velocity stays the same too!
I handed my physics notes from last year to a friend that's a year younger, forgetting that I'm probably going to need it lol. At least I know that I have some of the best resources (aka this youtube channel)
1. A ball is projected horizontally from the edge of the table that is 1.00 m high, and it strikes the floor at point 1.20 m from the base of the table. a. What is the initial speed of the ball? b. How high is the ball above the floor when its velocity vector makes a 45.0˚angle with the horizontal? please help in b
When the velocity vector makes a 45 degree with the horizontal, it means that the horizontal component and the vertical component of the velocity at that point is equal (tangent of 45 degrees is 1/1 =1). The velocity from part a is 2.67 m/s. This is always the horizontal component. At 45 degrees, this is also the vertical component. Use the kinematic equation using this value of the y-velocity, you get a height of 0.3637 meters from the top of the table. Since the table is 1 meter high, 1 minus 0.3637 equals 0.636 meters above the floor.
how would I find the angle at which the projectile (in the question I'm stuck with its a cannon ball hitting a ship) collides with the ground/object and the resultant velocity of the cannon ball hitting the ship?
You need to find vx and vy just when the projectile will hit the ground. vx=vox and vy=voy-gt. use the time when it hits the ground. Once you know both components of velocity you can use trigonometry to find the angle. tan(theta)=vy/vx and solve for the angle theta=atan(vy/vx). Hope this helps!
