<span>et us assume that the origin is the floor right below the 30 ft. fence
To work this one out, we'll start with acceleration and integrate our way up to position.
At the time that the player hits the ball, the only force in action is gravity where: a = g (vector)
ax = 0
ay = -g (let's assume that g = 32.8 ft/s^2. If you use a different value for gravity, change the numbers.
To get the velocity of the ball, we integrate the acceleration
vx = v0x = v0cos30 = 103.92
vy = -gt + v0y = -32.8t + v0sin40 = -32.8t + 60
To get the positioning, we integrate the speed.
x = v0cos30t + x0 = 103.92t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + 60t + 4
If the ball clears the fence, it means x = 0, y > 30
x = 0 -> 103.92 t - 350 = 0 -> t = 3.36 seconds
for t = 3.36s,
y = -16.4(3.36)^2 + 60*(3.36) + 4
= 20.45 ft
which is less than 30ft, so it means that the ball will NOT clear the fence.
Just for fun, let's check what the speed should have been :)
x = v0cos30t + x0 = v0cos30t - 350
y = 1/2*(-32.8)t² + v0sin30t + y0 = -16.4t² + v0sin30t + 4
x = 0 -> v0t = 350/cos30
y = 30 ->
-16.4t^2 + v0t(sin30) + 4 = 30
-16.4t^2 + 350sin30/cos30 = 26
t^2 = (26 - 350tan30)/-16.4
t = 3.2s
v0t = 350/cos30 -> v0 = 350/tcos30 = 123.34 ft/s
So he needed to hit the ball at at least 123.34 ft/s to clear the fence.
You're welcome, Thanks please :)
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Answer:
-4096
Step-by-step explanation:
8,-16,32,-64,128,-256,512,-1024,2048.-4096
Answer:
The values of x and y in the diagonals of the parallelogram are x=0 and y=5
Step-by-step explanation:
Given that ABCD is a parallelogram
And segment AC=4x+10
From the figure we have the diagonals AC=3x+y and BD=2x+y
By the property of parallelogram the diagonals are congruent
∴ we can equate the diagonals AC=BD
That is 3x+y=2x+y
3x+y-(2x+y)=2x+y-(2x+y)
3x+y-2x-y=2x+y-2x-y
x+0=0 ( by adding the like terms )
∴ x=0
Given that segment AC=4x+10
Substitute x=0 we have AC=4(0)+10
=0+10
=10
∴ AC=10
Now (3x+y)+(2x+y)=10
5x+2y=10
Substitute x=0, 5(0)+2y=10
2y=10

∴ y=5
∴ the values of x and y are x=0 and y=5
So $25 - $17.35 = $7.65
7.65÷ 17 = 45
45 = 45 min
595 is the number of at bats he would have in the season.