Depends on how strong you are. People can hit it anywhere between a few inches and 500 yards.
Answer:
The maximum height above the point of release is 11.653 m.
Explanation:
Given that,
Mass of block = 0.221 kg
Spring constant k = 5365 N/m
Distance x = 0.097 m
We need to calculate the height
Using stored energy in spring
...(I)
Using gravitational potential energy
....(II)
Using energy of conservation
Where, k = spring constant
m = mass of the block
x = distance
g = acceleration due to gravity
Put the value in the equation
Hence, The maximum height above the point of release is 11.653 m.
Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N
If you go to high you’ll run out of oxygen and possibly be blown off due to high winds.
