A ball is dropped from the top of a building. The height, , of the ball above the ground (in feet) is given as a function of time, , (in seconds) by
y = 1640 - 16t^2
y'= -32t
When does the ball hit the ground, and how fast is it going at that time? Give your answer in feet per second and in miles per hour (1 ft/sec = 15/22 mph).
v(t) = ______ mph
v(t) = ________ ft per second
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The trough is in the shape of a triangular prism. It is 5 ft long and its vertical cross sections are isosceles triangles with base 2ft and height 3ft. Water is being siphoned out of the trough at the rate of 2 cubic feet per minute. At any... time t, let h be the depth and V be the volumeof water in the trough.
To find when the ball hits the ground, we need to find the value of t when y = 0. So we can set the equation 1640 - 16t^2 = 0.
1640 - 16t^2 = 0
Divide both sides by 16:
t^2 = 102.5
Take the square root of both sides:
t = ±√102.5
Since we are looking for the time when the ball hits the ground, we only consider the positive square root:
t = √102.5
Now, let's find the velocity of the ball at that time. The velocity, v(t), is the derivative of the height function, y(t), with respect to time:
v(t) = y'(t) = -32t
Substituting t = √102.5:
v(t) = -32√102.5
To convert the velocity to miles per hour, we can multiply by the conversion factor 15/22:
v(t) = (-32√102.5) * (15/22) = -480√102.5/22
Therefore, the velocity of the ball at the time when it hits the ground is:
v(t) = -480√102.5/22 ft per second
To convert the velocity from ft per second to miles per hour:
v(t) = (-480√102.5/22) * (15/22)
Simplifying, we get:
v(t) ≈ -325.112 mph
So, the velocity of the ball at the time when it hits the ground is:
v(t) = -325.112 mph
and
v(t) = -480√102.5/22 ft per second
To find when the ball hits the ground, we need to solve the equation y = 0 because at the ground, the height will be 0.
Let's set y = 0
0 = 1640 - 16t^2
Now we need to solve for t. Rearranging the equation:
16t^2 = 1640
Dividing both sides by 16:
t^2 = 102.5
Taking the square root of both sides:
t = ±√102.5
Since time cannot be negative, we take the positive square root:
t ≈ 10.12 seconds
So, the ball hits the ground approximately 10.12 seconds after being dropped.
Now, let's find the speed at that time. The velocity, v(t), is the derivative of the height equation with respect to time (t).
v(t) = y' = -32t
Substituting the value of t when the ball hits the ground:
v(t) = -32 * 10.12
v(t) ≈ -323.84 ft per second
However, the velocity is negative because it is moving downward. To convert this to positive feet per second:
v(t) ≈ 323.84 ft per second
To convert feet per second to miles per hour, we can use the conversion factor 1 ft/sec = 15/22 mph.
v(t) ≈ (323.84 * 15/22) mph
v(t) ≈ 220.91 mph
So, the speed at the time the ball hits the ground is approximately 323.84 ft per second or 220.91 mph.