an arrow is shot straight up from a cliff 58.8 meters above the ground with an initial velocity of 49 meters per second
a. write an equation that represents this situation
b. Transform the equation to the format that can most easily be used to find the maximum height. What is the height?
c. Transform the equation to a different format to determine how many seconds it would take for the arrow to hit the ground.
d. Show how you would use the equation in part (a) to determine approximately when the arrow would be 100m above the base of the cliff
e. Show how you could determine the height of the arrow after 2 secs
f. Sketch the graph of the equation
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a. The equation that represents this situation can be written as:
h(t) = -9.8t^2 + 49t + 58.8
Where:
- h(t) represents the height of the arrow at time t.
- -9.8t^2 represents the effect of gravity on the arrow's height, with -9.8 being the acceleration due to gravity.
- 49t represents the initial upward velocity of the arrow, gradually decreasing due to gravity.
- 58.8 represents the initial height of the arrow from the ground.
b. To determine the maximum height, we need to find the vertex of the parabolic equation. The format that can most easily be used is the vertex form, given by:
h(t) = a(t - h)^2 + k
Where:
- a is the constant term in front of the squared term.
- (h, k) represents the coordinates of the vertex.
In our case, a = -9.8, and we need to find the coordinates of the vertex, which gives us:
h(t) = -9.8(t - h)^2 + k
To find h and k, we use the formulas:
h = -b / (2a) and k = f(h)
Using our equation, the maximum height is found at time t = h:
h = -49 / (2 * -9.8) = 2.5
Substituting this value back into the original equation, we can find the maximum height:
h(2.5) = -9.8(2.5)^2 + 49(2.5) + 58.8 = 108.05 meters
Therefore, the maximum height is 108.05 meters.
c. To determine the time it takes for the arrow to hit the ground, we need to find when the height equals zero. So, we set h(t) = 0 and solve for t:
0 = -9.8t^2 + 49t + 58.8
This is a quadratic equation, which can be solved using factoring, completing the square, or the quadratic formula.
d. To determine approximately when the arrow would be 100 meters above the base of the cliff, we need to find the time when the height equals 100. So, we set h(t) = 100 and solve for t:
100 = -9.8t^2 + 49t + 58.8
Again, this is a quadratic equation that needs to be solved.
e. To determine the height of the arrow after 2 seconds, we substitute t = 2 into the equation h(t):
h(2) = -9.8(2)^2 + 49(2) + 58.8
Solving this equation will give us the height of the arrow at 2 seconds.
f. The graph of the equation h(t) = -9.8t^2 + 49t + 58.8 will be a downward-opening parabola. The x-axis will represent time (t), and the y-axis will represent height (h). The vertex will be at the maximum height, and the parabola will intersect the x-axis at the time it hits the ground. Use the values calculated in parts b and c to plot the graph accurately.
a. To represent the situation of an arrow shot straight up from a cliff, we can use the equation of motion:
y = y₀ + v₀t - (1/2)gt²
where:
y represents the height of the arrow at time t,
y₀ represents the initial height (58.8 meters),
v₀ represents the initial velocity (49 meters per second),
g represents the acceleration due to gravity (approximately 9.8 meters per second squared), and
t represents time.
b. To find the maximum height, we set the velocity of the arrow at that point to zero. The equation becomes:
0 = v₀ - gt
Solving for t, we get:
t = v₀ / g
Substituting the values of v₀ and g, we have:
t = 49 m/s / 9.8 m/s² = 5 seconds
To find the maximum height, we can substitute this time (t = 5 seconds) into the equation of motion:
y = y₀ + v₀t - (1/2)gt²
y = 58.8 m + 49 m/s * 5 s - (1/2) * 9.8 m/s² * (5 s)²
y = 58.8 m + 245 m - (1/2) * 9.8 m/s² * 25 s²
y = 58.8 m + 245 m - 120.25 m
y = 183.55 m
So, the maximum height of the arrow is 183.55 meters.
c. To determine how many seconds it would take for the arrow to hit the ground, we set y (height) equal to zero in the equation of motion:
0 = y₀ + v₀t - (1/2)gt²
Rearranging the equation to solve for t, we get a quadratic equation:
(1/2)gt² - v₀t - y₀ = 0
Substituting the known values, the equation becomes:
(1/2) * 9.8 m/s² * t² - 49 m/s * t - 58.8 m = 0
This equation can be solved using various methods, such as factoring, quadratic formula, or graphical analysis. Solving the quadratic equation will give us the time it takes for the arrow to hit the ground.
d. To determine when the arrow would be 100 meters above the base of the cliff, we set y (height) equal to 100 meters in the equation of motion:
100 m = 58.8 m + 49 m/s * t - (1/2) * 9.8 m/s² * t²
Rearranging this equation, we get:
(1/2) * 9.8 m/s² * t² - 49 m/s * t - 58.8 m + 100 m = 0
Simplifying further, we have:
(1/2) * 9.8 m/s² * t² - 49 m/s * t + 41.2 m = 0
This quadratic equation can be solved to find the time when the arrow is approximately 100 meters above the base of the cliff.
e. To determine the height of the arrow after 2 seconds, we substitute t = 2 seconds into the equation of motion:
y = 58.8 m + 49 m/s * 2 s - (1/2) * 9.8 m/s² * (2 s)²
Simplifying this equation, we get:
y = 58.8 m + 98 m - (1/2) * 9.8 m/s² * 4 s²
y = 58.8 m + 98 m - 19.6 m
y = 137.2 m
So, the height of the arrow after 2 seconds is 137.2 meters.
f. The graph of the equation of motion is a parabolic curve representing the height of the arrow over time. The y-axis represents the height (y), and the x-axis represents time (t). At t = 0 seconds, the height is 58.8 meters (initial height), and then it reaches the maximum height of 183.55 meters at t = 5 seconds. The arrow hits the ground at some later time after that. The shape of the curve will be concave down because the acceleration due to gravity is downward.
a. V^2 = Vo^2 + 2g*hc.
hc = (Vf^2-Vo^2)/2g = Ht. above the cliff.
b. h max = ho + (Vf^2-Vo^2)/2g
h max = 58.8 + (0-49^2)/(-19.6) = 181.3 m. Above gnd.
c. V = Vo + g*Tr = 0.
49 -9.8Tr = 0, Tr = 5 s. = Rise time.
h max = 0.5g*Tf^2 = 181.3.
4.9Tf^2 = 181.3, Tf = 6.08 s. = Fall time.
Tr+Tf = 5 + 6.08 = 11.08 s. = Time to hit gnd.
d. V^2 = Vo^2 + 2g*h.
V^2 = 0 + 19.6(181.3-100) = 1593.48, V = 39.9 m/s.
V = Vo + g*t = 39.9 m/s.
0 + 9.8t = 39.9, t = 4.07 s.
e. h = Vo*t + 0.5g*t^2.
V0 = 49 m/s, t = 2 s., g = -9.8 m/s^2, h = ?.