A stone was thrown from the top of a cliff. The height, H meters, of the stone above sea level t seconds after it was released is given by
H=-5t+20t+60.
a)How high is the cliff? What makes you say this?
b) How high is the stone after two seconds?
c) What is the maximum height?
d) After how many seconds does it reach this maximum height?
but I think I see the problem
you mean
H = -5 t^2 + 20 t + 60
when t = 0, the height is 60. That is the height of the cliff
if t = 2
h = -5(4) + 20(2) + 60
= -20 +40 + 60 = 80
when dH/dt = 0 or vertical speed is 0 it stops going up
0 = -10 t + 20
t = 2 seconds to top
(that is answer d)
at t = 2
we did it already, 80 meters (part b = part c)
what your teacher did was say g =-10 m/s^2
F = m g = m a
so a = -10 m^s^2
then
v = Vi - g t = Vi - 10 t
h = Hi + Vi t - (10/2)t^2
Vi = 20
v = 0 at top = 20 - 10 t
so t at top = 2
h at top = 60 + 40 - 20 = 80
a) To find the height of the cliff, we need to consider the initial height of the stone when it was released. In this case, the equation given is H = -5t^2 + 20t + 60. The coefficient of t^2 term (-5) tells us that the stone was initially at a height of 0 meters. Therefore, the height of the cliff would be the constant term in the equation, which is 60 meters.
b) To find the height of the stone after two seconds, we substitute t = 2 into the equation H = -5t^2 + 20t + 60:
H = -5(2)^2 + 20(2) + 60
H = -5(4) + 40 + 60
H = -20 + 40 + 60
H = 80 meters
Therefore, the stone is 80 meters above sea level after two seconds.
c) The maximum height can be determined by finding the vertex of the parabolic equation H = -5t^2 + 20t + 60. The formula for the x-coordinate of the vertex of a quadratic equation in the form of ax^2 + bx + c is given by x = -b/2a.
For this equation, a = -5 and b = 20. Plugging these values into the formula, we get:
t = -20 / (2 * -5)
t = -20 / -10
t = 2
So, the maximum height occurs at t = 2 seconds.
To find the actual maximum height, substitute t = 2 into the equation:
H = -5(2)^2 + 20(2) + 60
H = -5(4) + 40 + 60
H = -20 + 40 + 60
H = 80 meters
Therefore, the maximum height of the stone is 80 meters.
d) We already determined that the stone reaches its maximum height at t = 2 seconds.
a) To determine the height of the cliff, we need to find the value of H when t is equal to zero. This is because the stone is released from the top of the cliff, so at t=0 seconds, the stone has not yet fallen and its height above sea level is solely determined by the height of the cliff.
Plugging t=0 into the equation H=-5t+20t+60, we get:
H = -5(0) + 20(0) + 60
H = 0 + 0 + 60
H = 60
Therefore, the height of the cliff is 60 meters since at the starting point (t=0), the stone is 60 meters above sea level.
b) To find the height of the stone after two seconds, we need to substitute t=2 into the equation H=-5t+20t+60.
Plugging in t=2, we get:
H = -5(2) + 20(2) + 60
H = -10 + 40 + 60
H = 90
So, the stone is at a height of 90 meters after two seconds.
c) The maximum height of the stone can be determined by observing the equation. In this case, H = -5t + 20t + 60 represents a quadratic function in the form of H(t) = at^2 + bt + c, where in our case a = 20, b = -5, and c = 60.
The maximum height is achieved when the quadratic function reaches its vertex. For a quadratic function in the form H(t) = at^2 + bt + c, the x-coordinate of the vertex is given by -b/2a.
In our case, -b = -(-5) = 5 and 2a = 2(20) = 40. Therefore, the x-coordinate of the vertex is -5/40 = -1/8.
To find the corresponding y-coordinate (maximum height), we can substitute -1/8 back into the equation H = -5t + 20t + 60:
H = -5(-1/8) + 20(-1/8) + 60
H = 5/8 - 5/2 + 60
H = 68.75
Hence, the maximum height of the stone is 68.75 meters.
d) To determine after how many seconds the stone reaches its maximum height, we can use the x-coordinate of the vertex, which we found earlier to be -1/8. Since t represents time in seconds, the negative value does not make logical sense in this context.
Therefore, we can conclude that the stone reaches its maximum height after 1/8 of a second or 0.125 seconds.