One gallon of paint covers an area of 25 m2. What is the thickness of the paint on the wall?
1 answer:
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Answer:
a) x = 40 t , y = 39 t , z = 6 + 32 t - 16 t ², b) x = 80 feet , y = 78 feet , the ball came into the field
Explanation:
a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis
Since the cast is in the center of the field, let's place the coordinate system
x₀ = 0
y₀ = 0
z₀ = 6 feet
x-axis (towards end zone, GOAL zone)
x = xo + v₀ₓ t
x = 40 t
y-axis (field width)
y = y₀ + t
y = 39 t
z axis (vertical)
z = z₀ + v_{oz} t - ½ g t²
z = 6 + 32 t - ½ 32 t²
z = 6 + 32 t - 16 t ²
b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive
z = 6
6 = 6 + 32 t - 16 t²
(t - 2)t = 0
t=0 s
t= 2 s
The ball position
x = 40 2
x = 80 feet
y = 39 2
y = 78 feet
the dimensions of the field from the coordinate system (center of the field) are
x_total = 150 feet
y _total = 80 feet
so we can see that the ball came into the field
(a) 0.40 s
First of all, let's find the initial speed at which Jordan jumps from the ground.
The maximum height is h = 1.35 m. We can use the following equation:
where
v = 0 is the velocity at the maximum height
u is the initial velocity
is the acceleration of gravity
Solving for u,
The time needed to reach the maximum height can now be found by using the equation
Solving for t,
Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of
h' = 1.35 - 0.20 = 1.15 m
Using again the equation
we find
And the corresponding time is
So the time to go from h' to h is
And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:
(b) 0.08 s
This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:
And the corresponding time is
So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2: