a diver springs from the edge of a cliff 80 ft above the ocean with an initial upward velocity of 8 ft/ s. How long will it take the diver to reach the water?
At a diving competition, divers will jump off a 64 foot high cliff. The diver's height above the water, at time t, is given by the equation . How long will it take the diver to hit the water and have a height of 0?
To calculate the time it takes for the diver to reach the water, we can use the equation of motion for free-falling objects:
h = ut + (1/2)gt^2
where:
h = height
u = initial velocity
g = acceleration due to gravity
t = time
In this case:
h = 80 ft (height of the cliff)
u = 8 ft/s (initial upward velocity)
g = 32 ft/s^2 (acceleration due to gravity, assuming downward direction)
Since the diver eventually reaches the water, the final height will be 0 ft. Therefore, we can rearrange the equation as follows:
0 = 80 + 8t - 16t^2
To solve this quadratic equation for t, we can set it equal to zero:
16t^2 - 8t - 80 = 0
Now we can solve this equation using a quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
where:
a = 16
b = -8
c = -80
Plugging the values into the formula:
t = (-(-8) ± √((-8)^2 - 4(16)(-80))) / (2 * 16)
Simplifying further:
t = (8 ± √(64 + 5120)) / 32
t = (8 ± √(5184)) / 32
t = (8 ± 72) / 32
Now we have two potential solutions for t:
t1 = (8 + 72) / 32 ≈ 2.5 seconds
t2 = (8 - 72) / 32 ≈ -1.9 seconds
Since time cannot be negative in this context, we discard the negative solution:
The diver will take approximately 2.5 seconds to reach the water.