gravitational potential is directly proportional to the height of the object relative to a reference line and is given as
PE = mgh
where m = mass of object , g = acceleration due to
gravity and h = height of the object above the reference line .
as the skydiver falls , its height above the ground decrease and hence the gravitational potential energy of the skydiver decrease.
as per conservation of energy , total energy of the skydiver must remain constant all the time . hence the decrease in potential energy appears as increase in kinetic energy by same amount to keep the total energy constant
KE + PE = Total energy
so as the skydiver falls , it gains speed and hence the kinetic energy of skydiver increase since kinetic energy is directly proportional to the square of the speed.
when the parachute opens, the skydiver experience force in upward which tries to balance the weight of the skydiver. hence the speed of the skydiver decrease until upward force becomes equal to the downward force. hence the kinetic energy decrease just after the parachute opens
= 3.456 × 1011
(scientific notation)
= 3.456e11
(scientific e notation)
= 345.6 × 109
(engineering notation)
(billion; prefix giga- (G))
= 345600000000
(real number)
Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
Answer:
32.46m/s
Explanation:
Hello,
To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2
A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

Where
Vf = final speed
Vo = Initial speed
=7.3m/S
A = g=acceleration
=9.81m/s^2
X = displacement
=51m}
solving for Vf

the speed with the ball hits the ground is 32.46m/s