A projectile is fired straight up in the air. Let h(t) be its height in meters above the ground t seconds after it was fired. Use the model
h(t) = −4.9t2 + 70t + 30
to answer the following questions. Feel free to use a calculator for computations, but show all your work for things like factoring or the quadratic formula. Use function notation when showing your work and answer the questions in words. Give decimal values for all answers, rounding to the nearest hundredth.
(a) What was the height of the projectile 8 seconds after it was fired?
(b) What was the height of the projectile a the moment when it was fired?
(c) At what time will the projectile hit the ground?
(d) What will the maximum height of the projectile be?
(e) Sketch the graph of h(t). Make it big and label all of the information you’ve found so far.
(f) What is the domain of the function?
( a )
Put t = 8 in your formula
h(8) = − 4.9 * 8 ^ 2 + 70 * 8 + 30
h(8) = − 4.9 * 64 + 560 + 30
h(8) = − 4.9 * 64 + 560 + 30
h(8) = − 313.6 + 590
h(8) = 276.4
( b )
Put t = 0 in your formula
h(0) = − 4.9 * 0 ^ 2 + 70 * 0 + 30
h(0) = − 4.9 * 0 + 0 + 30
h(0) = 0 + 30
h(0) = 30
( c )
When the projectile hit the ground then h = 0
So you must solve quadratic eguation:
− 4.9 t ^ 2 + 70 t + 30 = 0
Try that.
The solutions are:
t = ( 10 / 7 ) [ 5 - 2 sqroot ( 7 ) ] = - 0. 41643
and
t = ( 10 / 7 ) [ 5 + 2 sqroot ( 7 ) ] = 14.7021466
Time can't be negative so:
t = 14.70 sec
rounded on two decimal pieces
( d )
The parabola upside-down like in this case has a highest point when:
t = - b / 2 a
This point is called the "vertex".
In this case a = - 4.9 b = 70 c = 30 so:
t = - 70 / 2 * ( - 4.9 )
t = - 70 / - 9.8
t = 700 / 98
t = 2 * 2 * 25 * 7 / ( 2 * 7 * 7 )
t = 2 * 25 / 7
t = 50 / 7
t = 7.142857
t = 7.14 sec
rounded on two decimal pieces
Replace t = 50 / 7 in your equation.
hmax = t ( 50 / 7 )
hmax = - 4.9 * ( 50 / 7 ) ^ 2 + 70 * 50 / 7 + 30 =
- 4.9 * 250 / 49 + 7 * 10 * 50 / 7 + 30 =
- 4.9 * 2500 / ( 4.9 * 10 ) + 10 * 50 + 30 =
- 2500 / 10 + 500 + 30 =
- 250 + 500 + 30 =
250 + 30 = 280
( e )
Go on:
wolframalpha. c o m
When page be open in rectangle type:
plot − 4.9 t ^ 2 + 70 t + 30
Draw diagram on your paper then mark points.
( f )
In mathematical meaning the domain is - infinity to infinity
But time can't be negative and a projectile can't fly after hit the ground, so domain is t [ 0 , 14.70 ]
(a) To find the height of the projectile 8 seconds after it was fired, we substitute t = 8 into the given equation.
h(8) = -4.9(8)^2 + 70(8) + 30
= -4.9(64) + 560 + 30
= -313.6 + 590
= 276.4
Therefore, the height of the projectile 8 seconds after it was fired is 276.4 meters.
(b) To find the height of the projectile at the moment it was fired, we substitute t = 0 into the equation.
h(0) = -4.9(0)^2 + 70(0) + 30
= 0 + 0 + 30
= 30
Therefore, the height of the projectile at the moment it was fired is 30 meters.
(c) To find the time when the projectile hits the ground, we need to find when h(t) = 0.
-4.9t^2 + 70t + 30 = 0
This is a quadratic equation. We can either factor it or use the quadratic formula.
Factoring the equation gives us:
(t - 6)(-4.9t - 5) = 0
Setting each factor equal to zero gives us two possible solutions:
t - 6 = 0 or -4.9t - 5 = 0
Solving these equations, we find:
t = 6 or t = -1.02
Since time cannot be negative in this context, the projectile hits the ground after approximately 6 seconds.
Therefore, the projectile hits the ground at approximately 6 seconds.
(d) The maximum height of the projectile can be found by determining the vertex of the parabolic function h(t) = -4.9t^2 + 70t + 30.
The vertex is given by the formula: t = -b / (2a), where a = -4.9 and b = 70.
t = -70 / (2(-4.9))
= -70 / (-9.8)
= 7.14
Substituting t = 7.14 into the equation, we can find the maximum height.
h(7.14) = -4.9(7.14)^2 + 70(7.14) + 30
= -4.9(51.0996) + 499.8 + 30
= -249.958 + 499.8 + 30
= 279.842
Therefore, the maximum height of the projectile is approximately 279.842 meters.
(e) A sketch of the graph of h(t) is not possible in this text-based format. However, the graph is a downward-opening parabola, with the vertex at (7.14, 279.842). The x-axis represents time (t) and the y-axis represents height (h).
(f) The domain of the function is the set of all possible input values for t. In this case, since time cannot be negative in this context, the domain of the function is t ≥ 0.
(a) To find the height of the projectile 8 seconds after it was fired, we substitute t = 8 into the equation h(t) = -4.9t^2 + 70t + 30:
h(8) = -4.9(8)^2 + 70(8) + 30
= -4.9(64) + 560 + 30
= -313.6 + 560 + 30
= 276.4
Therefore, the height of the projectile 8 seconds after it was fired is 276.4 meters.
(b) To find the height of the projectile at the moment it was fired, we substitute t = 0 into the equation h(t) = -4.9t^2 + 70t + 30:
h(0) = -4.9(0)^2 + 70(0) + 30
= 30
Therefore, the height of the projectile at the moment it was fired is 30 meters.
(c) To find the time at which the projectile hits the ground, we need to find the value of t when h(t) = 0:
-4.9t^2 + 70t + 30 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -4.9, b = 70, and c = 30. Plugging these values into the quadratic formula:
t = (-70 ± √(70^2 - 4(-4.9)(30))) / 2(-4.9)
= (-70 ± √(4900 + 588)) / (-9.8)
= (-70 ± √(5488)) / (-9.8)
Using a calculator, we find that √(5488) ≈ 74.10. Plugging this value into the equation:
t = (-70 ± 74.10) / (-9.8)
We have two possible solutions:
t1 ≈ 1.57
t2 ≈ 12.54
Since time cannot be negative in this context, the projectile hits the ground approximately 12.54 seconds after it was fired.
(d) The maximum height of the projectile can be determined by finding the vertex of the quadratic function. The vertex of a quadratic in the form f(x) = ax^2 + bx + c is given by the formula:
x = -b / (2a)
In this case, a = -4.9 and b = 70. Plugging these values into the formula:
t = -70 / (2(-4.9))
= -70 / (-9.8)
= 7.14
To find the maximum height, we substitute t = 7.14 into the equation h(t):
h(7.14) = -4.9(7.14)^2 + 70(7.14) + 30
= -4.9(51.08) + 500 + 30
= -249.932 + 500 + 30
= 280.068
Therefore, the maximum height of the projectile is approximately 280.07 meters.
(e) To sketch the graph of h(t), we plot points using the values we have calculated. The graph should be a downward-facing parabola since the coefficient of the t^2 term is negative.
The vertex is located at (7.14, 280.07). We know the projectile hits the ground at approximately t = 12.54 and the height at t = 0 is 30 meters. We can also plot the point (8, 276.4).
Additionally, we can plot a few more points by substituting other values for t into the equation. For example, you can calculate h(4), h(10), and h(14). Plotting these points will help us to understand the shape of the parabola.
(f) The domain of the function h(t) is the set of all possible values for t. In this case, since the equation represents the height of a projectile above the ground, the domain would be all real numbers greater than or equal to 0, as time cannot be negative in this context. Thus, the domain of the function is t ≥ 0.