The height, s, of a ball thrown straight down with initial speed 32 ft/sec from a cliff 48 feet high is s(t) = -16t^2 - 32t + 48, where t is the time elapsed that the ball is in the air. What is the instantaneous velocity of the ball when it hits the ground?
To find the instantaneous velocity of the ball when it hits the ground, we need to find the time at which the height of the ball is 0.
We know that the height of the ball is given by the equation s(t) = -16t^2 - 32t + 48.
Setting s(t) = 0, we have -16t^2 - 32t + 48 = 0.
To solve this quadratic equation, we can either factor it or use the quadratic formula.
Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = -32, and c = 48.
Plugging in these values, we get:
t = (-(-32) ± √((-32)^2 - 4(-16)(48))) / (2(-16))
Simplifying further:
t = (32 ± √(1024 + 3072)) / (-32)
t = (32 ± √(4096)) / (-32)
t = (32 ± 64) / (-32)
We have two possible solutions:
1. t = (32 + 64) / (-32) = 96 / -32 = -3
2. t = (32 - 64) / (-32) = -32 / -32 = 1
Since time cannot be negative in this context, we disregard the first solution. Thus, the ball hits the ground at t = 1 second.
To find the instantaneous velocity at t = 1 second, we need to find the derivative of s(t) with respect to t:
v(t) = s'(t) = -32t - 32
Plugging in t = 1, we get:
v(1) = -32(1) - 32 = -32 - 32 = -64 ft/sec
Therefore, the instantaneous velocity of the ball when it hits the ground is -64 ft/sec.
To find the instantaneous velocity of the ball when it hits the ground, we need to determine the derivative of the position function, s(t), with respect to time, t. The derivative gives us the rate of change of the position function, which in this case represents the velocity.
The position function is given as s(t) = -16t^2 - 32t + 48. To find the derivative, we can apply the power rule and constant rule of differentiation.
Step 1: Apply the power rule
The power rule states that for a term of the form f(t) = t^n, the derivative is given by f'(t) = n*t^(n-1).
In our case, we have two terms involving t^2 and t, respectively.
Derivative of -16t^2: Using the power rule, we obtain -32t.
Derivative of -32t: Using the power rule, we obtain -32.
Step 2: Combine the derivatives of the individual terms
Since s(t) is the sum of the two terms, we can combine the derivatives obtained from Step 1.
s'(t) = -32t - 32.
The derivative s'(t) represents the velocity function. So, when the ball hits the ground, the time, t, will be the time at which s(t) equals zero. Therefore, we can find the velocity at that time by substituting t into the velocity function.
To find the time when the ball hits the ground, we set s(t) = 0 and solve for t.
0 = -16t^2 - 32t + 48
We can solve this quadratic equation by factoring or using the quadratic formula. For simplicity, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = -16, b = -32, and c = 48:
t = (-(-32) ± √((-32)^2 - 4(-16)(48))) / (2(-16))
= (32 ± √(1024 + 3072)) / (-32)
= (32 ± √4096) / (-32)
= (32 ± 64) / (-32)
We have two possible solutions:
1. t = (32 + 64) / (-32) = 96 / (-32) = -3
2. t = (32 - 64) / (-32) = -32 / (-32) = 1
Since time cannot be negative in this context, we discard the solution t = -3.
Therefore, the time at which the ball hits the ground is t = 1.
Now, we substitute t = 1 into the velocity function s'(t) to find the instantaneous velocity of the ball when it hits the ground.
s'(1) = -32(1) - 32
= -32 - 32
= -64 ft/sec.
So, the instantaneous velocity of the ball when it hits the ground is -64 ft/sec.
3thui45eht4in
well, what is t when s=0?
Use that value to calculate v(t) = -32t-32