A skydiver deploys his parachute when he is 1500 m directly above his desired la

Question

A skydiver deploys his parachute when he is 1500 m directly above his desired landing spot. He then falls through the air with a constant speed of 7.00 m/s . As he falls, there is a constant breeze blowing to west at 2.00 m/s .

At what standard angle does the parachutist move?

Use the standard convention for direction. On your picture, west should point to the left. Express your answer using three significant figures.

Part B

At what angle from the downward vertical does the parachutist move?

Part C

By what distance will the parachutist miss his desired landing spot?

Express your answer as a positive number using three significant figures.

Explanation / Answer

a. Vo = 7m/s at 270 degree + 2m/s at 180 degree.

X = 7*cos270 + 2*cos180 = -2 m/s.
Y = 7*sin270 + 2*sin180 = -7 m/s.

tanAr = Y/X = -7/-2 = 3.5
Ar = 74.05 = Reference angle.

A=180 + Ar = 180 + 74.05 = 254.05 in CCW.

A = 270 - 254.05 = 15.95 degree West of South.

Vo = -2/cos254.05 = 7.28 m/s at 254.05 degree
Xo = 7.28 * cos254.05 = -2 m/s.
Yo = 7.28 * sin254.05 = -7.0 m/s.

h = Vo*t + 0.5g*t^2 = 1500.
-7.28t + 4.9t^2 = 1500
4.9t^2 - 7.28t -1500 = 0
Use Quadratic formula:
Tf = 18.25 s. = Fall time.

Dx = Vx * Tf = -2m/s * 18.25=

Dx = 36.5

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.