1.
a.11/12=less than one
b. 77/72= greater than one
c.23/60= less than one
d. 31/56= less than one
2. <
2: 2/3; 3: 3/4; 4: 1/4; 5: 1/3; 6: 1/4; 7: 4/5; 8: 5/6-3/6=1/3
I think the answer is 56.81 seconds, the best time would just be the shortest time
<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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15 I may be wrong but If I am please provide more information.
