a diver dives from a cliff .when her centre of gravity is 46 ft above the surface of the water. her initial vertical velocity leaving the water is 9 ft per second. after how many secs does she hit the water? solve
You must use the equation h=-16t^2+vt+s
h=0
v=9
s=46
Then plug in and solve (I plugged the numbers into the equation but I'll let you solve it!) Hope this helps!
0=-16t^2+9t+46
Use the equation h=-16^2+vt+s
Plug in the numbers. In this case,
h=0
v=9
s=46
You will get 0=-16t^2+9t+46
(a) (b)(c)
Then use the quadratic formula and plug in the numbers to that.
1.475
So add 6 plus 09 and add -93
Well, let me dive into this question and calculate it for you.
We need to find the time it takes for the diver to hit the water. We can use the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height of the diver above the water (46 ft)
u = initial vertical velocity (9 ft/sec)
g = acceleration due to gravity (-32 ft/sec^2)
t = time it takes to hit the water (unknown)
To solve for t, let's substitute the given values into the equation:
46 = (9)t + (1/2)(-32)t^2
Now, let's simplify this equation and solve for t:
-16t^2 + 9t + 46 = 0
Hmm, this quadratic equation doesn't quite make a splash in the humor department, but let's continue.
Using the quadratic formula, we get:
t = (-9 ± √(9^2 - 4(-16)(46))) / (2(-16))
Calculating this equation, we find that t ≈ 3.2 seconds.
So, after approximately 3.2 seconds, the diver will hit the water. I hope she makes a big splash (but not too big).
To solve this problem, we can use the concept of projectile motion. We'll consider two phases: the upward motion and the downward motion.
In the upward motion:
1. We know the initial vertical velocity is 9 ft/s.
2. The acceleration due to gravity, which acts in the opposite direction of the motion, is -32 ft/s^2.
3. We want to find the time it takes for the diver to reach the highest point of her trajectory.
Using the kinematic equation: v = u + at, where:
- v is the final velocity,
- u is the initial velocity,
- a is the acceleration, and
- t is the time,
We can substitute the known values:
0 = 9 - 32t
Rearranging the equation:
32t = 9
t = 9/32 ≈ 0.28125 seconds (rounded to 5 decimal places)
Therefore, it takes about 0.28125 seconds for the diver to reach the highest point of her trajectory.
Now, for the downward motion:
1. We know the initial vertical velocity is 0 ft/s (at the highest point).
2. We want to find the time it takes for the diver to reach the surface of the water, given that her center of gravity is 46 ft above it.
Using the same kinematic equation:
v = u + at
Substituting the known values:
0 = 0 + (-32)t
Simplifying the equation:
-32t = 0
t = 0 seconds
In the downward motion, the time is 0 seconds because the diver starts falling immediately after reaching the highest point.
Therefore, after 0.28125 seconds, the diver hits the water.