<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:
a) False
b) False
c) True
d) False
e) False
Step-by-step explanation:
a. A single vector by itself is linearly dependent. False
If v = 0 then the only scalar c such that cv = 0 is c = 0. Hence, 1vl is linearly independent. A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, only a single zero vector is linearly dependent, while any set consisting of a single nonzero vector is linearly independent.
b. If H= Span{b1,....bp}, then {b1,...bp} is a basis for H. False
A sets forms a basis for vector space, only if it is linearly independent and spans the space. The fact that it is a spanning set alone is not sufficient enough to form a basis.
c. The columns of an invertible n × n matrix form a basis for Rⁿ. True
If a matrix is invertible, then its columns are linearly independent and every row has a pivot element. The columns, can therefore, form a basis for Rⁿ.
d. In some cases, the linear dependence relations among the columns of a matrix can be affected by certain elementary row operations on the matrix. False
Row operations can not affect linear dependence among the columns of a matrix.
e. A basis is a spanning set that is as large as possible. False
A basis is not a large spanning set. A basis is the smallest spanning set.
Answer:
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Job B because every year he gets a $2800 every year and if he continues for 5 years he will get a total of $14000 but if he works at job A he gets $5000 after 5 years
