A circus performer is juggling clubs while standing on stilts. He released a club from a point 8 feet above the ground with an initial vertical velocity of 10 ft per second.
a. Write the equation that models the height [h] (in feet) of the club as a function of the time [t] in seconds.
b.How high does the club go? (Round answer to nearest tenth)
c. How long after the club is released does it reach its maximum height? (Round to nearest tenth)
d. When is the club at the height of 9 feet?
e. After the performer releases the club, how much time does he have to catch the club on its way down when the club reaches a height of 7 feet? (Round answers to nearest tenth)
f. If the performer does not catch the club, when will it land on the ground? Round your answer to the nearest tenth.
a. h = ho + Vo*t -16t^2.
ho = Initial height.
Vo = Initial velocity.
b. h = ho + (Vf^2 - Vo^2) / 2g,
h = 8 + (0 - (10)^2) / -64.4 = 9.55 Ft.
above ground.
c. t = (Vf - Vo) / g,
t = (0 - 10) / -32.2 = 0.31s.
d. h = 8 + 10t - 16t^2 = 9 Ft.
-16t^2 + 10t + 8 = 9,
-16t^2 + 10t - 1 = 0,
Use Quadratic Formula to find t:
t = 0.125s.
e. d = Vo*t + 0.5gt^2 = 9.55 - 7 = 2.55 Ft.
0 + 16t^2 = 2.55,
t^2 = 0.159,
t(dn) = 0.40s.
T = t(up) + t(dn) = 0.31 + 0.40 = 0.7s to catch the club.
f. d = Vo*t + 16t^2 = 9.55 Ft.
0 + 16t^2 = 9.55,
t^2 =0.597,
t = 0.8s.
0.8 seconds
a. The equation that models the height [h] (in feet) of the club as a function of time [t] in seconds is given by:
h(t) = -16t^2 + 10t + 8
b. To find the maximum height, we need to find the vertex of the quadratic equation. The formula for the x-coordinate of the vertex is given by -b/2a, where a = -16 and b = 10. Plugging in the values, we get:
t = -10/(2*(-16))
t = -10/(-32)
t = 5/16
To find the height at this time, we can substitute t = 5/16 into the equation:
h(5/16) = -16(5/16)^2 + 10(5/16) + 8
h(5/16) = -16(25/256) + 50/16 + 8
h(5/16) = -25/16 + 50/16 + 128/16
h(5/16) = 153/16
So, the maximum height the club reaches is 9.6 feet.
c. The time it takes for the club to reach its maximum height is equal to t = 5/16, which is approximately 0.313 seconds.
d. To find when the club is at a height of 9 feet, we can set h(t) = 9 and solve for t:
9 = -16t^2 + 10t + 8
-16t^2 + 10t + 8 - 9 = 0
-16t^2 + 10t - 1 = 0
Solving this quadratic equation using the quadratic formula, we get:
t = (-b ± √(b^2 - 4ac))/(2a)
t = (-10 ± √(10^2 - 4*(-16)*(-1)))/(2*(-16))
t = (-10 ± √(100 - 64))/(-32)
t = (-10 ± √(36))/(-32)
t = (-10 ± 6)/(-32)
The two possible solutions are:
t1 = ( -10 + 6 ) / ( -32 ) = 1 / 8 ≈ 0.125
t2 = ( -10 - 6 ) / ( -32 ) = -16 / -32 = 1 / 2 ≈ 0.5
So, the club is at a height of 9 feet at approximately 0.125 seconds and 0.5 seconds.
e. To find how much time the performer has to catch the club when it reaches a height of 7 feet, we need to solve for t when h(t) = 7:
7 = -16t^2 + 10t + 8
-16t^2 + 10t + 1 = 0
Solving this quadratic equation using the quadratic formula, we get:
t = (-b ± √(b^2 - 4ac))/(2a)
t = (-10 ± √(10^2 - 4*(-16)*1))/(2*(-16))
t = (-10 ± √(100 + 64))/(-32)
t = (-10 ± √(164))/(-32)
Since we are interested in the time it takes for the club to reach a height of 7 feet on its way down, we only consider the positive square root:
t = (-10 + √(164))/(-32)
Using a calculator, we find that t ≈ 0.056 seconds.
f. To find when the club lands on the ground, we need to solve for t when h(t) = 0:
0 = -16t^2 + 10t + 8
Solving this quadratic equation using the quadratic formula, we get:
t = (-b ± √(b^2 - 4ac))/(2a)
t = (-10 ± √(10^2 - 4*(-16)*8))/(2*(-16))
t = (-10 ± √(100 + 512))/(-32)
t = (-10 ± √(612))/(-32)
Since we are interested in when the club lands on the ground, we only consider the positive square root:
t = (-10 + √(612))/(-32)
Using a calculator, we find that t ≈ 1.474 seconds.
To answer these questions, we need to analyze the motion of the club as it is released and moves upward, reaches its maximum height, and then falls back down to the ground. We'll use some kinematic equations to model the height of the club and solve for various quantities.
a. The equation that models the height [h] of the club as a function of time [t] can be determined using the kinematic equation:
h(t) = h0 + v0t - (1/2)gt^2
where h0 is the initial height (8 feet), v0 is the initial vertical velocity (10 ft/s), g is the acceleration due to gravity (negative 32 ft/s^2).
Therefore, the equation is:
h(t) = 8 + 10t - (1/2)(32)t^2
b. To find the maximum height reached by the club, we need to determine when the vertical velocity becomes zero. The club reaches its maximum height at the peak of its trajectory, where the change in velocity changes direction from upwards to downwards.
The vertical velocity equation is given by:
v(t) = v0 - gt
Setting v(t) = 0, we can solve for t to find the time at which the club reaches its maximum height.
0 = 10 - 32t
Solving for t, we get:
t = 10/32 ≈ 0.3125 seconds
Now we can substitute this value of t into the height equation to find the maximum height:
h(t) = 8 + 10(0.3125) - (1/2)(32)(0.3125)^2
h(t) ≈ 8.5 feet
So, the club reaches a maximum height of approximately 8.5 feet.
c. We have already determined that the club reaches its maximum height at t ≈ 0.3125 seconds.
d. To find when the club is at a height of 9 feet, we can set the height equation equal to 9 and solve for t:
9 = 8 + 10t - (1/2)(32)t^2
Rearranging and simplifying:
16t^2 - 20t - 8 = 0
Using the quadratic formula, we find two possible values for t:
t ≈ -0.2857... and t ≈ 1.0357...
Since time cannot be negative in this context, we discard the negative value.
Therefore, the club is at a height of 9 feet at t ≈ 1.0357 seconds.
e. To determine how much time the performer has to catch the club when it reaches a height of 7 feet, we need to find the time t when the height h(t) = 7. We can use the height equation and solve for t:
7 = 8 + 10t - (1/2)(32)t^2
16t^2 - 20t + 6 = 0
Using the quadratic formula, we find two possible values for t:
t ≈ 0.2609... and t ≈ 1.7391...
We are interested in the time t after the club is released, so we discard the smaller value.
Therefore, the performer has approximately 1.7391 seconds to catch the club when it reaches a height of 7 feet.
f. To find when the club lands on the ground, we can find the time t when the height h(t) = 0. We can again use the height equation and solve for t:
0 = 8 + 10t - (1/2)(32)t^2
16t^2 - 20t - 8 = 0
Using the quadratic formula, we find two possible values for t:
t ≈ -0.2857... and t ≈ 1.0357...
Since time cannot be negative in this context, we discard the negative value.
Therefore, the club lands on the ground at approximately t ≈ 1.0357 seconds.