Ruth wants to purchase a new video game that costs $59.00 and several bags of chips, c, that sell for $1.99 each. She can spend
no more than $75.00. Which inequality and its solution represents her possible purchase?
2 answers:
You might be interested in
Answer:
A. h = h₀ + u·t - 1/2·g·t²
B. 30.9 m
C. 4.73 seconds
2. A. h = h₀ + u·t - 1/2·a·t²
B. 67.8 m
C. Approximately 25.73 seconds
3. A. h = h₀ + u·t - 1/2·g·t²
B. 31.38 m
C. Approximately 5.142 seconds
D. Approximately 32.9 m
Step-by-step explanation:
The given parameters are;
The initial height of the ball, h₀ = 15 m
The upward velocity with which the ball is thrown, u = 20 m/sec.
The acceleration due to gravity, g = 9.8 m/s²
A. The relation between the height, h, and the time, t, after the ball is released is given as follows;
h = h₀ + u·t - 1/2·g·t²
B. The height of the ball after 3 seconds is given by substitution as follows;
At t = 3 seconds, h = 15 + 20 × 3 - 1/2 × 9.8 × 3² = 30.9
The height of the ball, h, after 3 seconds is h = 30.9 m
C. The time the ball takes to hit the ground = 2 × The time it takes to maximum height + The time it takes the ball to fall with an initial velocity of 20 m/sec for 15 m height
The time it takes to maximum height, , is given as follows;
v = u - g·
Where;
v = The final velocity = 0 at maximum height
Therefore, we have;
0 = 20 - 9.8 ×
∴ = 20/9.8 ≈ 2.0408
The time it takes to maximum height, ≈ 2.0408 seconds
The time it takes the ball to fall with an initial velocity of 20 m/sec for 15 m height, is given as follows;
v₂² = u₂² + 2·g·h₀
v₂² = 20² + 2×9.8×15 = 694
v₂ = √694 ≈ 26.344 m/s
v₂ ≈ 26.344 m/s
From, v₂ = u₂ + g·, we have;
26.344 = 20 + 9.8×t
9.8· = 26.344 - 20 = 6.344
∴ = 6.344/9.8 ≈ 0.647
The ball will hit the ground after 2 × +
≈ 2 × 2.0408 + 0.647 ≈ 4.7286
The ball will hit the ground after approximately 4.7286 ≈ 4.73 seconds
2. When the ball is thrown upward from the Moon, we have;
The acceleration due to gravity on the moon, a = 1.6 m/s², therefore, we have;
A. The relation between the height, h, and the time, t, after the ball is released is given as follows;
h = h₀ + u·t - 1/2·a·t²
B. The height of the ball after 3 seconds is given by substitution as follows;
At t = 3 seconds, h = 15 + 20 × 3 - 1/2 × 1.6 × 3² = 67.8
The height of the ball, h, thrown on the Moon, after 3 seconds is h = 67.8 m
C. The time the ball takes to hit the ground = 2 × The time it takes to maximum height + The time it takes the ball to fall with an initial velocity of 20 m/sec for 15 m height
The time it takes to maximum height, , is given as follows;
v = u - a·
Where;
v = The final velocity = 0 at maximum height
Therefore, we have;
0 = 20 - 1.6 ×
∴ = 20/1.6 = 12.5
The time it takes to maximum height, = 12.5 seconds
The time it takes the ball to fall with an initial velocity of 20 m/sec for 15 m height, is given as follows;
v₂² = u₂² + 2·a·h₀
v₂² = 20² + 2×1.6×15 = 448
v₂ = √448 ≈ 21.166 m/s
v₂ ≈ 21.166 m/s
From, v₂ = u₂ + a·, we have;
21.166 = 20 + 9.8×t
1.6· = 21.166 - 20 = 1.166
∴ = 1.166/1.6 ≈ 0.72785
The ball will hit the ground after 2 × +
≈ 2 × 12.5 + 0.72875 = 25.72875 ≈ 27.73
The ball will hit the ground after approximately 25.73 seconds
3. The height from which the ball is kicked, h₀ = 1 m
The initial velocity of the ball, u = 25 m/sec
The acceleration due to gravity, g = 9.8 m/s²
The relationship between the height, h and the time, t after the ball is released, is given as follows;
h = h₀ + u·t - 1/2·g·t²
B. The height of the ball after 2 seconds is given as follows;
At t = 2, h = 1 + 25 × 2 - 1/2 × 9.81 × 2² = 31.38
The height of the ball, after 2 seconds, h = 31.38 m
C. The time it takes the ball to hit the ground is given by the following kinematic equation, as follows;
h = h₀ + u·t - 1/2·g·t²
At the ground level, h = 0, therefore, we have;
0 = 1 + 25·t - 4.9·t²
Therefore, by the quadratic formula, we have;
t = (-25 ± √(25² - 4×(-4.9)×1))/(2 × -4.9)
Therefore, t ≈ 5.142, or t ≈ -0.03969
Given that the time is a natural number, we have, t ≈ 5.142 seconds
D. The maximum height, the ball reaches is given as follows;
From the kinematic equation, v² = u² - 2·g·h,
Where;
v = 0 at maximum height
h = The height the ball reaches above the initial height, we have;
0² = u² - 2·g·h
u² = 2·g·h
h = u²/(2·g) = 25²/(2 × 9.8) ≈ 31.888
= h₀ + h = 1 + 31.888 ≈ 32.9
The maximum height the ball reaches, ≈ 32.9 m
To translate 4 units to the right, you replace 
with 
So we get
So we get