Suppose we are standing on top of a 300-ft tower and we are holding a shiny new penny.We do not hurt it down, for that would be dangerous and wrong! We toss it up at a velocity of 10 feet per second and then when it comes down it just happens to plummet to the ground below
At a time t seconds after the toss, the velocity is given by v(t)=-32t+10.Its distance from the sidewalk, in feet, is given by p(t)=-16t^2+1-t+300
a. At what time is the penny's velocity zero?
b. When is the penny at its highest point?
c. What is the maximum height achieved by the penny?
d. How long is te penny in the air?
e.How fast is it going when it hits the ground?
(a) -32t+10 = 0
(b) see (a) above
(c) use (a) in p(t)
(d) solve p(t)=0
(e) use (d) in v=-32t+10
The penny's height at any time, t, seconds would be given by the equation h(t)=-16t²+10t+300. It reaches its' maximum height at -10/-32, or 0.3125 seconds (-b/2a). This gives it a maximum height of 301.5625 ft. The penny lands at h(t)=0, or 16t²-10t-300=0; so total time in the air would be 4.6539 seconds. Its' velocity would be -32(4.6539)+10, or -138.9248 ft/sec downward
a. To find the time when the penny's velocity is zero, we need to set v(t) = 0, since velocity is given by v(t) = -32t + 10.
-32t + 10 = 0
-32t = -10
t = -10 / -32
t ≈ 0.3125 seconds
Therefore, the penny's velocity is zero at approximately 0.3125 seconds.
b. To find when the penny is at its highest point, we need to find the vertex of the parabolic function p(t) = -16t^2 + 1t + 300, since distance is given by p(t).
The vertex of a quadratic function in the form of ax^2 + bx + c is given by x = -b / 2a.
In this case, a = -16 and b = 1.
t = -1 / 2(-16)
t = 1 / 32
t ≈ 0.03125 seconds
Therefore, the penny is at its highest point at approximately 0.03125 seconds.
c. The maximum height achieved by the penny is the value of p(t) at the time t = 0.03125 seconds.
p(0.03125) = -16(0.03125)^2 + 0.03125 + 300
p(0.03125) ≈ 300.09766 feet
Therefore, the maximum height achieved by the penny is approximately 300.09766 feet.
d. The time the penny is in the air can be determined by finding the duration from when it is tossed up to when it hits the ground. To find this, we need to solve for t when p(t) = 0.
-16t^2 - t + 300 = 0
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / 2a, with a = -16, b = -1, and c = 300.
t = (1 ± √((-1)^2 - 4(-16)(300))) / (2(-16))
t = (1 ± √(1 + 19200)) / (-32)
t ≈ (1 ± √19201) / -32
The positive value of t represents the time the penny is in the air. Without calculating the approximate value, we can determine that t > 0.
Therefore, the penny is in the air for some positive amount of time.
e. The speed at which the penny is going when it hits the ground can be determined by finding the velocity at the time t from part d.
We can determine this by evaluating v(t) = -32t + 10 at the time t when the penny hits the ground.
Since we determined that t > 0 in part d, we can calculate the velocity when it hits the ground by substituting t with the positive value we obtained.
v(t) = -32t + 10
Therefore, the velocity of the penny when it hits the ground is given by v(t) = -32t + 10.
To answer these questions, we need to analyze the equations given for velocity and distance. Here's how we can find the answers:
a. To find the time when the penny's velocity is zero, we set v(t) = 0 and solve for t:
-32t + 10 = 0
-32t = -10
t = -10 / -32
t = 5/16 seconds
b. To find when the penny is at its highest point, we need to find the maximum value of the distance function p(t). We can use calculus to find the vertex of the parabola representing the distance equation.
The vertex of a parabola in the form of p(t) = at^2 + bt + c is given by t = -b / (2a). In this case, a = -16, b = -1, and c = 300.
So, t = -(-1) / (2(-16))
t = 1 / 32 seconds
c. The maximum height achieved by the penny is the value of p(t) at the highest point, which can be found by substituting the value of t from the previous step into the distance equation:
p(t) = -16t^2 + 1 - t + 300
p(1/32) = -16(1/32)^2 + 1 - (1/32) + 300
p(1/32) = 301.9375 feet
d. The time the penny is in the air can be calculated by finding the time it takes for the distance to reach zero. This occurs when p(t) = 0, so we can solve for t:
-16t^2 + 1 - t + 300 = 0
-16t^2 - t + 301 = 0
Using the quadratic formula, t = (-b ± sqrt(b^2 - 4ac)) / (2a)
t = (-(-1) ± sqrt((-1)^2 - 4(-16)(301))) / (2(-16))
Solve and select the positive value of t, which gives the time the penny is in the air.
e. To find how fast the penny is going when it hits the ground, we can use the velocity equation v(t) = -32t + 10 and substitute the value of t found in part d. This will give us the speed of the penny when it hits the ground.