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
for the first part solve
0 = 1640 - 16t^2
16t^2 = 1640
t^2 = 102.5
t = 10.124 seconds
sub that into y' to get the velocity in ft/sec
Use the formula given to you to change to mph
To find when the ball hits the ground, we need to solve for t when the height, y, is equal to zero. The equation for the height of the ball is given as:
y = 1640 - 16t^2
Setting y = 0, we have:
0 = 1640 - 16t^2
Rearranging the equation, we get:
16t^2 = 1640
Dividing both sides by 16, we have:
t^2 = 102.5
Taking the square root of both sides, we find:
t = ±√102.5
Since time cannot be negative in this context, we can ignore the negative square root value. Thus, the ball hits the ground at:
t = √102.5
To find how fast the ball is going at that time, we need to find the velocity, v(t), which is the derivative of the height function y with respect to time t:
v(t) = y'(t) = -32t
Substituting the value of t when the ball hits the ground:
v(√102.5) = -32(√102.5)
Now, we need to convert the velocity from feet per second to miles per hour. Given that 1 ft/sec = 15/22 mph, we can calculate:
v(√102.5) = -32(√102.5) * (15/22)
Simplifying the expression, we get:
v(√102.5) ≈ -42.17 ft/sec
To convert the velocity to miles per hour:
v(√102.5) ≈ -42.17 ft/sec * (15/22) = -28.64 mph
Therefore, the ball hits the ground at approximately √102.5 seconds and is going approximately -42.17 ft/sec or -28.64 mph at that time.