Question

A 580-g squirrel with a surface area of 850 cm2 falls from a 4.0-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the cross-sectional area of the squirrel can be approximated as a rectangle of width 11.0 cm and length 22 cm. Note, the squirrel may not reach terminal velocity by the time it hits the ground. Give the squirrel's terminal velocity, not it's velocity as it hits the ground.)

What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?

Answer 1

its terminal velocity is 19.70 m/s

the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s

Step-by-step explanation:

Firstly,

given that

m = 580g = 0.58kg

Area A = 0.11 * 0.22 = 0.0242m

g = 9.8

idensity constant p = 1.21 kg/m^3

the terminal velocity of the sphere Vt is ;

Vt = √ ( 2mg / pCA)

we substitute

Vt = √ ( (2*0.58*9.8) / (1.21*1*0.0242)

Vt = √ (11.368 / 0.029282)

Vt = √ ( 388.22)

Vt = 19.70 m/s

its terminal velocity is 19.70 m/s

What will be the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance?

The Velocity of the person is;

V2 = √ 2ax

V2 = √ ( 2 * 9.8 * 4 )

V2 = √ (78.4)

V2 = 8.85 m/s

the velocity of a 58.0-kg person hitting the ground, assuming no drag contribution in such a short distance is 8.85 m/s