Miss E. deWater, the former platform diver of the Ringling Brothers' Circus, dives from a 80.0-meter high platform into a shallow bucket of water (see diagram at right). Determine her speed and her height after each second of fall.
When t = 0 s: speed = 0 m/s and height = 80 m
Should I use kinematic equations to find acceleration then plug that in?
A CAR TRAVELLING AT 2OMS ACCELERATION UNIFORMLY AT 3MS^2 CACULATE IT VELOCITY AFTER 3OS KM/HR
Yes, you can use kinematic equations to find the acceleration and then use it to determine Miss E. deWater's speed and height after each second of fall. The following steps will guide you through the calculations:
1. Determine the acceleration of Miss E. deWater using the kinematic equation: Δy = vi * Δt + 0.5 * a * (Δt)^2, where Δy is the change in height, vi is the initial velocity (0 m/s in this case), Δt is the time interval (1 s), and a is the acceleration.
Rearranging the equation to solve for acceleration:
a = (Δy - vi * Δt) / (0.5 * (Δt)^2)
Since Δy = -80 m (negative sign indicates the decreasing height), substitute the given values:
a = (-80 m - 0 m/s * 1 s) / (0.5 * (1 s)^2)
2. Calculate the acceleration:
a = (-80 m) / (0.5 s^2) = -160 m/s^2
3. Determine the speed after each second using the equation: vf = vi + a * t, where vf is the final velocity, vi is the initial velocity (0 m/s in this case), a is the acceleration (-160 m/s^2), and t is the time interval (1 s).
For t = 1 s:
vf = 0 m/s + (-160 m/s^2) * 1 s = -160 m/s
4. Determine the height after each second using the equation: Δy = vi * Δt + 0.5 * a * (Δt)^2, where Δy is the change in height, vi is the initial velocity (0 m/s in this case), a is the acceleration (-160 m/s^2), and Δt is the time interval (1 s).
For t = 1 s:
Δy = 0 m/s * 1 s + 0.5 * (-160 m/s^2) * (1 s)^2 = -80 m
Therefore, after 1 second of fall, Miss E. deWater's speed is -160 m/s (negative sign indicates downward velocity), and her height is -80 m (negative sign indicates decreasing height from the initial 80 m).
Yes, you can use the kinematic equations to find the acceleration and then use that information to calculate the speed and height after each second of the fall.
To begin, let's determine the acceleration of the platform diver. We can use the equation:
v = u + at
where:
- v is the final velocity (speed)
- u is the initial velocity (which is 0 in this case)
- a is the acceleration
- t is the time
Since the initial velocity is 0 m/s, the equation simplifies to:
v = at
Rearranging the equation, we have:
a = v / t
Now we need to find the acceleration. In this scenario, the only force acting on Miss E. deWater is gravity, so we can use the acceleration due to gravity, which is approximately 9.8 m/s^2.
Using the equation:
a = 9.8 m/s^2
Now, let's calculate the speed and height after each second:
At t = 1 s:
- Acceleration (a) = 9.8 m/s^2
- Time (t) = 1 s
Using the equation:
v = at
v = 9.8 m/s^2 * 1 s
v = 9.8 m/s
At t = 1 s, the speed of Miss E. deWater is 9.8 m/s.
Now, to find the height after 1 second, we can use the equation of motion:
s = ut + (1/2)at^2
where:
- s is the displacement (height in this case)
- u is the initial velocity (0 m/s)
- t is the time
- a is the acceleration (9.8 m/s^2)
Using the equation:
s = (0 m/s) * 1 s + (1/2)(9.8 m/s^2)(1 s)^2
s = 0 m + (1/2)(9.8 m/s^2)(1 s)^2
s = 0 m + (1/2)(9.8 m/s^2)(1 s)
s = 0 m + 4.9 m = 4.9 m
After 1 second of falling, Miss E. deWater's speed is 9.8 m/s and her height is 4.9 meters.
You can repeat this process for subsequent seconds to find the speed and height after 2 seconds, 3 seconds, and so on.
V = Vo + g*T.
Vo = 0.
g = 10 m/s^2.
h = ho - 0.5g*Tf^2 = 0.
5Tf^2 = 80.
Tf^2 = 16.
Tf = 4 s. = Fall time.
Use T in the range of 0 to 4 seconds.
(T,V,h).
(0,0,80).
(1,10,75).
(2,20,60).
(3,30,35).
(4,40,0).