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Mmmmmm yummy oh yeah um mmmm I’m so high
Number 3 is gonna be 2 and Number 4 will be 294 I hope this would help u a little bit I will figure out the Other questions
Assuming this little game of catch took place on planet earth, negative acceleration due to gravity is -9.8 m/s .
Converting 30ft/s to m/s, initial velocity was 30ft/s x 0.305ft/meter = 9.14m/s
Let's find how long it took for the velocity to equal zero, meaning when the ball reached it's highest point and, for a split second, stopped in mid-air before falling back down.
V(t) = Vi + a*t , where V(t) is velocity as a function of time, a is acceleration due to gravity, and t is time. Set V(t) = 0
0 = 9.14 + (-9.8)* t Add -9.8t to both sides
9.8t = 9.14 Divide both sides by 9.8
t = 0.93 seconds
Let's say your hand is the base point, or where h=zero. We want to find how high above your hand the ball went before it started coming down. Using the distance, or in this case height, formula:
h = Vi*t + (1/2)at² Plug in Vi, a, and our t value, 0.93
h= 9.14 * 0.93 + (1/2)(9.8)(0.93²)
h= 8.5 + 4.9 (0.865)
h = 8.5+ 4.27
h = 12.74 meters
The ball made it 12.74 meters above your hand. Your friends had was one foot above yours, so let's subtract .305 meters to see how far it dropped from the peak height to his hand.
12.74-.305 = 12.43 meters
Let's use the distance formula again to see how long it took to come down. Remember that this time, initial velocity is zero, since the ball starts off suspended in the air.
-12.43 = 0*t + (1/2)(-9.8)(t²) Divide both sides by -9.8/2, or -4.9
2.5374 = t²
t = 1.59
The ball took .93 seconds to go up, and 1.59 seconds to come down to your friend's glove. The total time the ball was in the air:
.93 + 1.59 = 2.52 seconds
Answer:
y = 13*( -x/9 + 1/5)
Step-by-step explanation:
Given:
- The curve has an equation as follows:

Find:
a. Verify that the given point (2,2) lies on the curve.
b. Determine an equation of the line tangent to the curve at the given point.
Solution:
- To verify whether the point lies on the given curve we will substitute the coordinates of the point into the equation as follows:
44 = 5*(2)^2 + 3*(2)(2) + 3*(2)^2
44 = 20 + 12 + 12
44 = 44 ......Hence proven.
- The equation of the line tangent to the curve is expressed as a linear function as follows:
y = m*x + C
Where, m is the gradient of the line.
C is the y-intercept.
m = Δy / Δx = dy/dx
- We will take the derivative of the given curve with respect to x as follows:

- Evaluate y' at the point (2,2) we get:
y' = - ( 10(2) + 3(2) ) / ( 3(2) + 6(2) )
y' = - ( 26 ) / (18)
y'= m = - 13/9
- To evaluate C, we will use the point (2,2) for linear expression above with m as follows:
y = -13*x/9 + C
2 =-13*(2)/9 + C
C = 13 / 5
- The equation of the tangent is as follows:
y = 13*( -x/9 + 1/5)