Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.
Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m
Now you need the final speed to use it as initial speed of the next part.
Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s
Part B) Free fall
Maximum height, y max ==> Vf = 0
Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s
ymax = yo + Vo*t - g[t^2] / 2
ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m
Answer: ymax = 10084.2m
Answer:
x=2d
Explanation:
initial stretch in the spring is d
so using Hook's law
at equilibrium position
k×d=mg
where k= spring constant
m= mass of fish
g= acceleration due to gravity.
d=mg/k ................ (1)
in second case by energy conservation
1/2 kx^2=mgx
x=2mg/k
using equation 1
x=2d
Answer:
520000 or 520000 pa
Force = 520N
Area of contact = 0.001
Pressure: 520000 or 520000
Answer:
0 (there is no speed)
Explanation:
If an object is at rest, it is not moving, and it doesn't have a speed, so the speed is zero.
