The height of an object dropped from a 144-foot-tall building is given by the function s=−16t2+144 , where s is in feet and t is in seconds. Find the object’s speed when it hits the ground.
To find the speed of the object when it hits the ground, we need to find the velocity at that moment. Velocity is the derivative of displacement with respect to time.
The displacement function s = -16t^2 + 144 is a quadratic equation, so we can differentiate it to find the velocity function.
Differentiating s with respect to t:
ds/dt = -32t
To find the velocity at the moment the object hits the ground, we need to find the value of t when s = 0 (since the object hits the ground when the displacement is 0).
Setting s = 0 in the displacement function:
-16t^2 + 144 = 0
16t^2 = 144
t^2 = 144/16
t^2 = 9
t = ±√9
t = ±3
We discard the negative value of t because time cannot be negative in this context.
At t = 3 seconds, the object hits the ground.
To find the velocity at t = 3 seconds, substitute this value of t into the velocity function:
v = ds/dt
v = -32(3)
v = -96 ft/s
The object's speed when it hits the ground is 96 ft/s.
To find the object's speed when it hits the ground, we need to find the time it takes for the object to reach the ground. The object hits the ground when its height, s, is equal to zero.
So, we set the equation -16t^2 + 144 = 0
To solve this equation, we can factor out a common factor of 16:
16(-t^2 + 9) = 0
Now, we set each factor equal to zero:
-t^2 + 9 = 0
Rearranging the equation, we have:
t^2 - 9 = 0
This equation is a difference of squares, so we can factor it as:
(t + 3)(t - 3) = 0
Setting each factor equal to zero, we get:
t + 3 = 0 or t - 3 = 0
Solving for t, we have:
t = -3 or t = 3
Since time cannot be negative, we discard the negative value of t. Therefore, the object takes 3 seconds to reach the ground.
Now, to determine the object's speed when it hits the ground, we need to find the derivative of the function s(t) with respect to t, which gives us the velocity function v(t). The derivative of -16t^2 + 144 is -32t.
So, v(t) = -32t
Substituting t = 3 into the velocity function, we get:
v(3) = -32(3)
v(3) = -96
Therefore, the object's speed when it hits the ground is -96 feet per second. Note that the negative sign indicates that the object is moving downward.
To find the object's speed when it hits the ground, we need to determine the value of t when s, the height, is zero. This is because the object hits the ground when the height is zero.
Given the function s = -16t^2 + 144, we can set s equal to zero and solve for t:
0 = -16t^2 + 144
Now, we need to solve this quadratic equation for t. To do this, we can factor out a common factor of 16:
0 = 16(-t^2 + 9)
Next, divide both sides by 16:
0 = -t^2 + 9
Since we want to solve for t, let's isolate the t^2 term by subtracting 9 from both sides:
t^2 = 9
Now, take the square root of both sides to solve for t:
t = ±√9
t = ±3
Since we are interested in the time when the object hits the ground, we only consider the positive value of t:
t = 3
The object hits the ground 3 seconds after being dropped.
To find the object's speed when it hits the ground, we need to find the derivative of the position function with respect to time (velocity). Taking the derivative of -16t^2 + 144 with respect to t gives us:
v = d/dt(-16t^2 + 144)
v = -32t
Now, substitute the value of t when the object hits the ground, which is t = 3:
v = -32(3)
v = -96
Therefore, the object's speed when it hits the ground is -96 feet per second (considering that the negative sign indicates the direction of motion, which is downward in this case).