An object is projected vertically upwards. It’s height hm at time t seconds is given by
h=20t-3/2t^2 - 2/3t^3, find:
a) the time at which it is momentarily at rest
b) correct to 2dp, the maximum height reached by the object?
To find the time at which the object is momentarily at rest, we need to find the time when the velocity of the object is zero. The velocity can be obtained by differentiating the height equation with respect to time.
a) Time at which the object is momentarily at rest:
To find the time when the velocity is zero, let's differentiate the height equation, h, with respect to time, t.
h(t) = 20t - (3/2)t^2 - (2/3)t^3
Differentiate both sides of the equation:
dh/dt = 20 - 3t - 2t^2
Now, set dh/dt equal to zero and solve for t:
20 - 3t - 2t^2 = 0
We have a quadratic equation. To solve it, you can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Applying this formula to our equation, we have:
t = (-(-3) ± √((-3)^2 - 4 * (-2) * 20)) / (2 * (-2))
Simplifying further:
t = (3 ± √(9 + 160)) / (-4)
t = (3 ± √169) / (-4)
t = (3 ± 13) / (-4)
So, we have two possible solutions:
t = (3 + 13) / (-4) = 16 / (-4) = -4 (discard since time cannot be negative)
t = (3 - 13) / (-4) = -10 / (-4) = 2.5
Thus, the time at which the object is momentarily at rest is 2.5 seconds.
b) Maximum height reached by the object:
To find the maximum height, we need to find the vertex of the parabolic equation. The vertex of the parabola represents the maximum point.
We can use the equation for the vertex of a parabola: t = -b / (2a)
In our case, a = -2/3, b = -3/2, and c = 20. So, substituting these values into the equation:
t = -(-3/2) / (2 * (-2/3))
t = (3/2) / (4/3)
t = (3/2) * (3/4)
t = 9/8 = 1.125
Now, substitute this value of t back into the height equation to find the maximum height:
h(t) = 20t - (3/2)t^2 - (2/3)t^3
h(1.125) = 20(1.125) - (3/2)(1.125)^2 - (2/3)(1.125)^3
Simplifying this expression will give you the maximum height reached by the object.
Note: Make sure to calculate the result accurately to 2 decimal places for the final answer.
To find the time at which the object is momentarily at rest, we need to find the time when the velocity is equal to zero. The velocity can be found by taking the derivative of the height function with respect to time.
a) Find the derivative of the height function h(t):
h'(t) = 20 - 3t^2 - 2t^2
To find the time when the velocity is zero, we set h'(t) equal to zero and solve for t:
0 = 20 - 3t^2 - 2t^3
Simplifying the equation:
2t^3 + 3t^2 - 20 = 0
This is a cubic equation that needs to be solved for t. Unfortunately, solving cubic equations does not have a simple step-by-step process. You can use numerical methods or a graphing calculator to find the solutions for t.
b) To find the maximum height reached by the object, we need to find the maximum point on the height function h(t).
To find the maximum point, we need to find the value of t where the derivative h'(t) changes from positive to negative. This indicates a change from increasing to decreasing height.
Using the derivative h'(t) from above:
h'(t) = 20 - 3t^2 - 2t^2
Set h'(t) equal to zero:
0 = 20 - 3t^2 - 2t^2
Combine like terms:
0 = 20 - 5t^2
Rearrange the equation:
5t^2 = 20
Divide both sides by 5:
t^2 = 4
Take the square root of both sides:
t = ± 2
Since time cannot be negative in this context, we take the positive solution:
t = 2 seconds
Substitute t = 2 back into the height function to find the maximum height:
h(2) = 20(2) - 3/2(2^2) - 2/3(2^3)
h(2) = 40 - 3(2) - 2(8/3)
h(2) = 40 - 6 - 16/3
h(2) = 40 - 6 - 5.33
h(2) = 34 - 5.33
h(2) ≈ 28.67
Therefore, the maximum height reached by the object is approximately 28.67 units.