You toss a rock up vertically at an initial speed of 55 feet per second and release it at an initial height of 6 feet. The rock will remain in the air for_______ seconds.
It will reach a maximum height of_______ feet after ________ seconds.
To find the time the rock remains in the air, we can use the kinematic equation:
h = h0 + v0t - (1/2)gt^2
Where:
h = final height (0 feet, as it reaches back to the ground)
h0 = initial height (6 feet)
v0 = initial velocity (55 feet/second, upward)
g = gravitational acceleration (32 feet/second^2, downward)
t = time
Setting h = 0, h0 = 6, v0 = 55, and g = 32, we have:
0 = 6 + 55t - (1/2)32t^2
Rearranging the equation:
16t^2 - 55t - 12 = 0
We can solve this quadratic equation to find the time the rock remains in the air. Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where:
a = 16
b = -55
c = -12
Calculating:
t = (-(-55) ± √((-55)^2 - 4 * 16 * -12)) / (2 * 16)
t = (55 ± √(3025 + 768)) / 32
t = (55 ± √(3793)) / 32
Therefore, the rock will remain in the air for approximately 3.03 seconds or 0.47 seconds (rounded to two decimal places).
To find the maximum height, we can use the formula for vertical displacement:
y = y0 + v0t + (1/2)gt^2
Since we are interested in the maximum height, we need to find the time at the peak of the trajectory. The time at the peak can be found using the formula:
v = v0 + gt
At the peak, the velocity (v) is zero, so we can rearrange the equation:
0 = 55 + (32)(t_peak)
Solving for t_peak:
t_peak = -55/32
Substituting this value back into the displacement formula:
y = y0 + v0(-55/32) + (1/2)(32)(-55/32)^2
y = 6 - (55)(55/32)(1/2)
y ≈ 407.03 feet (rounded to two decimal places)
The rock reaches a maximum height of approximately 407.03 feet after approximately 0.47 seconds (rounded to two decimal places).
To find the time the rock will remain in the air, you can use the equation of motion for vertical displacement:
𝑑𝑦 = 𝑣0𝑦𝑡 + 0.5𝑔𝑡^2
Where:
𝑑𝑦 is the vertical displacement (in this case, the difference in height between the initial and final positions)
𝑣0𝑦 is the initial vertical velocity (in this case, 55 feet per second)
𝑔 is the acceleration due to gravity (approximately 32.2 feet per second squared)
𝑡 is the time (in seconds)
Since the vertical displacement is the same as the initial height (6 feet) when the rock falls back to its original position, we can set 𝑑𝑦 = 6 and solve for 𝑡.
6 = 55𝑡 - 0.5(32.2)𝑡^2
Simplifying and rearranging the equation:
0.5(32.2)𝑡^2 - 55𝑡 + 6 = 0
This is a quadratic equation and can be solved using the quadratic formula:
𝑡 = (-𝑏 ± √(𝑏^2 - 4𝑎𝑐)) / (2𝑎)
In this case, 𝑎 = 0.5(32.2), 𝑏 = -55, and 𝑐 = 6.
Calculating the values:
𝑡 = (-(-55) ± √((-55)^2 - 4(0.5(32.2))(6))) / (2(0.5(32.2)))
Simplifying further:
𝑡 = (55 ± √(3025 - 96.8)) / (32.2)
𝑡 = (55 ± √(2928.2)) / 32.2
Using a calculator, we find two possible solutions for 𝑡:
𝑡≈ 1.007 seconds or 𝑡≈ 6.974 seconds
Therefore, the rock will remain in the air for approximately 1.007 seconds or 6.974 seconds.
To find the maximum height reached by the rock, we can use the equation:
𝑣𝑦 = 𝑣0𝑦 + 𝑔𝑡
At the maximum height, the vertical velocity (𝑣𝑦) will be 0. We can rearrange and solve for 𝑡:
0 = 55 - 32.2𝑡
Solving for 𝑡:
32.2𝑡 = 55
𝑡 = 55 / 32.2
Calculating the value:
𝑡 ≈ 1.707 seconds
Substituting the value of 𝑡 back into the equation for 𝑑𝑦:
𝑑𝑦 = 𝑣0𝑦𝑡 + 0.5𝑔𝑡^2
𝑑𝑦 = 55(1.707) + 0.5(32.2)(1.707)^2
Calculating the value:
𝑑𝑦 ≈ 48.140 feet
Therefore, the rock will reach a maximum height of approximately 48.140 feet after 1.707 seconds.
from the given data:
h = -16t^2 + 55t + 6
this is a parabola, we need its vertex.
the t of the vertex is -b/(2a) = -55/-32
= 55/32 seconds
plug that into equation to find max height
it is in the air for h > 0
let's see when h = 0
0 = -16t^2 + 55t + 6
16t^2 - 55t - 6 = 0
solve using the formula, since this does not factor.
ignore the negative root, just use the positive one
so the rock will be in the air for
0 > t > (above positive answer for t)