A rescue plane wants to drop supplies to isolated mountain climbers on a rocky r

Question

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4 m/s), how far in advance of the recipients ( horizontal distance) must the goods be dropped ? A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4 m/s), how far in advance of the recipients ( horizontal distance) must the goods be dropped ?

Explanation / Answer

Given

Velocity, vx = 250 kmph = 69.44 m/s

Distance, yo = 235 m

vyo = 0, ay = - g, y = 0 m

(a) From Newton's 2nd equation of motion

y = yo +  vyo t + (1 / 2) ay t2

   0 = yo + 0 + (1 / 2) a t2

   t = (- 2 yo / a) = 6.93 s

So, horizontal distance, x = vx t = 69.44 x 6.93 = 480.89 m

Please rate my answer if you find it helpful, good luck...

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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