How do you find the horizontal velocity if you know the initial velocity?

Answer 1

The way your question is phrases, numerous correct answers exist.
BUT, I am going to make the assumption that you are looking for the horizontal velocity of a projectile lauched with an initial velocity of v at an angle theta to the horizontal in a uniform downward gravitational field.
You can divide the initial launch velocity of the projectile into its vertical and horizontal compoents if one know the launch angle using trigonometric functions.
V_x = v * cos (theta)
V_y = v * sin (theta)
where V_x and V_y are the initial horizontal and vertical velocities of the projectile repectively when launched at an angle theta above the horizontal with an intial velocity of v.
The horizontal velocity of the object (neglecting air resistance) can be treated as a constant, unlike the vertical velocity which will reach a maximum as the object experiences a constant downward acceleration due to gravity.

Answer 2

You have to take the original direction of the velocity into account, and find the component of the velocity in the horizontal direction, usually using trigonometry.
Ex. V(i) cos X = V(h), where V(i) is initial velocity, V(h) horizontal velocity and X = angle above horizontal.

Answer 3

If the initial velocity "V" is at angle b to the horizontal direction, then the horizontal velocity is V cos b.

Answer 4

Here's in which you start. Write out the entire equations you'll want, then fill within the bits you understand. The predominant factor to keep in mind is that you've got one set of equations for the vertical movement, and one other set for the horizontal movement. So you're employed with just one size at a time. What permits you to attach what is taking place in a single dimention with teh different is time vertical acceleration = -nine.eight vertical speed = -nine.8t + v v0 vertical role = -four.nine(t^two) + v v0t +pv0 horizontal speed = vh0 horizontal role = vh0t + ph0 So, your inital vertical speed, or v v0 is zero. Your are most effective kicking ahead, now not up or down. We'll make your preliminary vertical role, or pv0 7.nine. Your inital horizontal speed , vh0 is what we're watching for, and we will make your horizontal role, ph0 -15. So the zero,zero factor is solely in which the rock enters the cave close its ceiling. So we fill in: vertical speed = -nine.8t vertical postion = -four.nine(t^two) + 7.nine horizontal speed = vh horizontal role = vht - 15 So we clear up for t in which the rock reaches the right vertical role of zero. Then we clear up for the vh in which, for that t, the rock has the correct horizontal role, additionally of zero. (You might have additionally made the zero,zero factor in which the rock begins, then your inital role values are zero and zero, and then you definitely clear up for in which the rock's coordinates have been 7.nine,-15. The math works out the equal.) Once you've got the correct horizontal speed, see how lengthy it takes the rock to journey the horizontal distance to the again wall, then see how a long way the rock falls vertically in that point.

Answer 5

without air resistence the horizontal velocity is constant is V0* cos b

Answer 6

YOU STUDY YOUR CLASS NOTES, YOUR TEXT BOOKS and then YOU DO YOUR HOMEWORK