If a football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees, then the vertical component of the initial velocity would be 12.65 m/s
<h3>What is Velocity?</h3>
The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second.
As given in the problem A football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees
The horizontal component of the velocity is given by
Vx = Vcosθ
The vertical component of the velocity is given by
Vy = Vsinθ
As we have to find the vertical component of the velocity given that speed of 16.71 m/s at an angle of 49.21 degrees from the ground
Vy = 16.71 × sin49.21°
Vy = 12.65 m/s
Thus, the vertical component of the velocity would be 12.65 m/s
Learn more about Velocity from here
brainly.com/question/18084516
#SPJ1
Answer:
f = 5 cm
Explanation:
using the thin lens equation, given as follows:
where,
f = focal length = ?
do = the distance of object from lens = 20 cm
di = the distance of image from lens = 6.6667 cm
Therefore,
<u>f = 5 cm</u>
It should be A.
A ball bouncing is moving so if it’s moving that means it has kinetic energy. It also has potential energy because when it hits the floor it kind of stops so it has potential.
-Hope this helps.
We actually don't need to know how far he/she is standing from the net, as we know that the ball reaches its maximum height (vertex) at the net. At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m. So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY....... We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION: v² = u² + 2as 0 = u² - 2gh u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity) So rearranging, velocity = (velocity in y direction only) / sin 3° = √(2gh)/sin 3° = (√(2 x 9.8 x 0.33)) / sin 3° = 49 m/s at 3° to the horizontal (2 sig figs)
