The horizontal motion has no effect on the vertical drop.
From a drop, the distance the ball falls in 'T' seconds is
D = 4.9 T^2
so
2.2 = 4.9 T^2
T^2 = 2.2/4.9
T^2 = 0.449 sec^2
T = 0.67 second
Answer:
Miter joint
Explanation:
Made by fastening together usually perpendicular parts with the ends cut at an angle
The wind speed is 20
the plane is 430mph in still air
Answer:
<h2>Velocity</h2>
Explanation:
When a body is launched in air and allowed to fall freely under the influence of gravity, the motion experienced by the body is known as a projectile motion. The body is launched at a particular velocity and at an angle theta to the horizontal. The velocity of the body ca be resolved towards the horizontal component and the vertical component.
Along the horizontal Ux = Ucos(theta)
Along the vertical Uy = Ucos(theta)
Ux and Uy are the velocities of the body along the horizontal and vertical components respectively.
This means that the only factor connecting horizontal and vertical components of projectile motion is <u>its velocity</u> since we are able to calculate the velocity of the body along both components irrespective of its initial velocity.
<span>(6.0x10^-22, -1.40x10^-21, 0) kg*m/s
Momentum is a conserved quantity. The total momentum of the system before and after the interactions will not change. So, let's look at the momentum before the interaction.
(3.2x10^-21, 0, 0) kg*m/s and (0,0,0) kg*m/s
After the interaction
(2.6x10^-21, 1.40x10^-21, 0) kg*m/s
and the other proton has to have a momentum that when added to this momentum equal the original value. Since the y and z vectors were initially 0, all we need for the y and x vector values of the result is to negate them. The x vector value will be
3.2x10^-21 - 2.6x10^-21 = 0.6x10^21 = 6.0x10^-22. So the other proton will have a momentum of
(6.0x10^-22, -1.40x10^-21, 0) kg*m/s</span>