Answer:
and 
Explanation:
The first sentence of this question is not explanatory enough. However, I'll assume the force to be 15N
to the horizontal
Required
Solve for the x and y components
Since the given angle is to the horizontal, the x and y coordinates are calculated using the following illustrations.
---- y component
---- x component
Calculating the y component.
Substitute 15 for Force and 36.7 for 
becomes
Make y the subject
Calculating the x component.
Substitute 15 for Force and 36.7 for
becomes
Make y the subject
<em>Hence, the x and y components of the force are: 8.964N and 12.027N respectively.</em>
We have that the time in seconds, minutes, and hours is
From the Question we are told that
Velocity 
Distance 
Generally the equation for the Time is mathematically given as
Therefore
And
For more information on this visit
brainly.com/question/12319416?referrer=searchResults
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)
Answer:
9.6 Ns
Explanation:
Note: From newton's second law of motion,
Impulse = change in momentum
I = m(v-u).................. Equation 1
Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.
Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)
Substitute into equation 1
I = 2.4[2.5-(-1.5)]
I = 2.4(2.5+1.5)
I = 2.4(4)
I = 9.6 Ns
