A frustrated office worker throws his laptop out of his business’s high-rise
office building. The height, h, in meters of the laptop above the ground at t
seconds can be modelled by 2 h(t) 125 5t .
Estimate the instantaneous rate at which the laptop hits the ground.
sorry the equation is h(t)=125=125-5t^2
h(t) = 125 - 5t^2
Take the derivative to find the velocity:
v(t) = dh(t)/dt = -10 t
Now:
Find the time when height is 0.
Find the velocity at that time.
but when h is 0 t=0....
h = 125-5t^2 = 0
-5t^2 + 125 = 0
t^2 - 25 = 0
(t+5)(t-5) = 0
t = -5
t = 5 s.
V = g*t = 9.8 * 5 = 49 m/s.
To estimate the instantaneous rate at which the laptop hits the ground, we need to calculate the derivative of the height function with respect to time. In this case, the height function is given by h(t) = 2t^2 + 125t.
To find the derivative, we need to apply the power rule for derivatives. The power rule states that if we have a term of the form ax^n, where a is a constant and n is a real number, then the derivative is given by (d/dx)(ax^n) = anx^(n-1).
Applying the power rule, we differentiate each term separately:
(d/dt)(2t^2) = 4t
(d/dt)(125t) = 125
Therefore, the derivative of the height function h(t) = 2t^2 + 125t is given by h'(t) = 4t + 125.
Now, to estimate the instantaneous rate at which the laptop hits the ground, we need to evaluate the derivative at the time when the laptop hits the ground, which is when the height is zero. So, we set h(t) = 0 and solve for t:
2t^2 + 125t = 0
Factoring out t, we get:
t(2t + 125) = 0
This gives us two possible values for t: t = 0 (which corresponds to the initial time when the laptop was thrown) and t = -125/2. Since time cannot be negative in this case, we can discard t = -125/2.
Therefore, the time at which the laptop hits the ground is t = 0, which is the initial time when the laptop was thrown.
Next, we evaluate the derivative at t = 0:
h'(0) = 4(0) + 125 = 125
So, the estimated instantaneous rate at which the laptop hits the ground is 125 meters per second.