Joshua jumps from a platform diving board that is 32 feet above the water. The position of the diver can be modeled by s(t) = -16t2+16t+32, where s is his position and t is the seconds that have gone by since he jumped.
When does the diver hit the water, and what was his velocity?
Oh good grief, feet :( old text
had initial speed 16 ft/s up ? I guess ?
h = 32 + 16 t - 16 t^2
0 = 32 + 16 t - 16 t^2
solve quadratic
t = [ -16 +/- sqrt ( 256+2048) ] /-32
t = 2 seconds
v = dh/dt = 16 - 32 t
= 16 - 64
= -48 ft/s
To find when the diver hits the water, we need to determine the value of t when s(t) equals zero. This is because at the time when the diver hits the water, his position is at the water level.
Let's set s(t) = 0:
-16t^2 + 16t + 32 = 0
Now, we can solve this quadratic equation. We can either factor it or use the quadratic formula. To use the quadratic formula, which is a = -16, b = 16, and c = 32, the formula is:
t = (-b ± √(b^2 - 4ac)) / (2a)
Applying the formula, we have:
t = (-(16) ± √((16)^2 - 4(-16)(32))) / (2(-16))
t = (-16 ± √(256 + 2048)) / (-32)
t = (-16 ± √(2304)) / (-32)
t = (-16 ± 48) / (-32)
This gives us two possible solutions:
1. When t = (-16 + 48) / (-32) = 32 / -32 = -1
2. When t = (-16 - 48) / (-32) = -64 / -32 = 2
Since time cannot be negative in this context, the diver hits the water at t = 2 seconds.
To find the velocity of the diver at that time t = 2, we can calculate the derivative of s(t) with respect to t. The derivative of s(t) is the velocity function v(t).
v(t) = s'(t) = -32t + 16
Substituting t = 2 into this equation gives us:
v(2) = -32(2) + 16
v(2) = -64 + 16
v(2) = -48
Therefore, the velocity of the diver when he hits the water is -48 feet per second.
To find out when the diver hits the water and what his velocity is, we need to solve the equation s(t) = -16t^2 + 16t + 32 for t.
To do this, set s(t) equal to zero since the water level is at height zero.
0 = -16t^2 + 16t + 32
Now we have a quadratic equation. We can solve it by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 16, and c = 32. Plug in these values:
t = (-16 ± √(16^2 - 4(-16)(32))) / (2(-16))
Simplifying further:
t = (-16 ± √(256 + 2048)) / (-32)
t = (-16 ± √2304) / (-32)
t = (-16 ± 48) / (-32)
Now we get two solutions:
t1 = (-16 + 48) / (-32)
t2 = (-16 - 48) / (-32)
Calculating further:
t1 = 32 / -32
t1 = -1
t2 = -64 / -32
t2 = 2
Therefore, the diver hits the water at t = -1 (one second before jumping) and t = 2 (two seconds after jumping).
To find the velocity at these times, we can take the derivative of the position function, s(t), with respect to time, t:
v(t) = s'(t)
v(t) = -32t + 16
Now we can substitute t = -1 and t = 2 into the velocity function to find the diver's velocities at those times:
v(-1) = -32(-1) + 16
v(-1) = 32 + 16
v(-1) = 48 ft/s
v(2) = -32(2) + 16
v(2) = -64 + 16
v(2) = -48 ft/s
Therefore, the diver's velocity at t = -1 is 48 ft/s and at t = 2 is -48 ft/s. The negative velocity at t = 2 indicates that the diver is moving downward at that time.