Multi-step prob During a trampoline competition, a trampolinist leaves the mat when her center of gravity is 6 feet above the ground. She has an initial vertical velocity of 32 ft/sec.
a. Use the vertical motion model (h=-16t^2+vt+s) to write an eqn that models the height h (in ft) of the center of gravity of the trampolinist as a function of the time t (in seconds) into her jump.
b. Does her center of gravity reach a height of 24 ft during the jump? Of so, at what time (s)?
c. On another jump, the trampolinist leaves the mat when her center of gravity is 6 ft above the ground and with an initial vertical velocity of 35 ft/sec. Does her center of gravity reach a hight of 24 ft on this jump? If so, at what time (s)?
(a) well? It's obviously
h = -16t^2 + 32t + 6
(b) solve for t when h=24
(c) same , but use 35t instead of 32t
a. The equation that models the height h (in ft) of the center of gravity of the trampolinist as a function of time t (in seconds) into her jump is:
h = -16t^2 + vt + s
Where:
h = height of the center of gravity (in ft)
t = time (in seconds)
v = initial vertical velocity (in ft/sec)
s = initial height (in ft)
In this case, the initial height (s) is given as 6 ft and the initial vertical velocity (v) is given as 32 ft/sec. Therefore, the equation becomes:
h = -16t^2 + 32t + 6
b. To find out if her center of gravity reaches a height of 24 ft during the jump, we need to solve the equation when h = 24 ft. We can set up the equation as follows:
24 = -16t^2 + 32t + 6
Simplifying the equation:
-16t^2 + 32t + 6 - 24 = 0
-16t^2 + 32t - 18 = 0
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 32, and c = -18. Plugging in these values:
t = (-32 ± √(32^2 - 4(-16)(-18))) / (2(-16))
t = (-32 ± √(1024 - 1152)) / (-32)
t = (-32 ± √(-128)) / (-32)
Since we have a negative value inside the square root, the equation has no real solutions. Therefore, her center of gravity does not reach a height of 24 ft during the jump.
c. Similarly, for the second jump with an initial vertical velocity of 35 ft/sec, we need to set up the equation:
h = -16t^2 + 35t + 6
To find out if her center of gravity reaches a height of 24 ft, we set h = 24:
24 = -16t^2 + 35t + 6
Simplifying the equation:
-16t^2 + 35t + 6 - 24 = 0
-16t^2 + 35t - 18 = 0
Using the quadratic formula:
t = (-35 ± √(35^2 - 4(-16)(-18))) / (2(-16))
t = (-35 ± √(1225 - 1152)) / (-32)
t = (-35 ± √(73)) / (-32)
We have real solutions in this case, so we can solve for t. By calculating the values inside the square root, we get:
t = (-35 ± √(73)) / (-32)
t ≈ -0.0456 or t ≈ 2.0456
Since time cannot be negative, the trampolinist's center of gravity reaches a height of 24 ft at approximately t = 2.0456 seconds on this jump.
a. To write an equation that models the height h of the trampolinist's center of gravity as a function of time t, we will use the given information from the vertical motion model:
h = -16t^2 + vt + s
In this equation:
- h represents the height of the trampolinist's center of gravity (in feet).
- t represents the time (in seconds) into the jump.
- v represents the initial vertical velocity (in feet per second).
- s represents the initial height (in feet) above the ground.
Since the trampolinist leaves the mat when her center of gravity is 6 feet above the ground, we can set s = 6 in the equation.
Therefore, the equation that models the height of the trampolinist's center of gravity is:
h = -16t^2 + 32t + 6
b. To determine if the trampolinist's center of gravity reaches a height of 24 feet during the jump, we need to substitute h = 24 into the equation and solve for t.
24 = -16t^2 + 32t + 6
Rearranging the equation, we get:
16t^2 - 32t + 18 = 0
This is a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 16, b = -32, and c = 18.
Plugging in these values, we get:
t = (-(-32) ± √((-32)^2 - 4(16)(18))) / (2(16))
= (32 ± √(1024 - 1152)) / 32
= (32 ± √(-128)) / 32
Since the expression inside the square root is negative, it means that there are no real solutions for t. Hence, the trampolinist's center of gravity does not reach a height of 24 feet during the jump.
c. Let's repeat the same process for the second jump. The equation that models the height of the trampolinist's center of gravity is:
h = -16t^2 + 35t + 6
Substituting h = 24, we get:
24 = -16t^2 + 35t + 6
Rearranging the equation, we get:
16t^2 - 35t + 18 = 0
Using the quadratic formula, we have:
t = (-(-35) ± √((-35)^2 - 4(16)(18))) / (2(16))
= (35 ± √(1225 - 1152)) / 32
= (35 ± √73) / 32
We have real solutions for t in this case. To find the exact values, we can evaluate the expressions separately:
t = (35 + √73) / 32 ≈ 1.59 seconds
t = (35 - √73) / 32 ≈ 0.22 seconds
Therefore, the trampolinist's center of gravity reaches a height of 24 feet during the second jump at approximately 0.22 seconds.