Question

A World War II-era dive bomber is being used to drop food and supplies to outposts on remote mountaintops. During one such drop, the bomber descends with velocity v directed at an angle of 36.1° below the horizontal. The cargo is released at an altitude of 2430 m and reaches its intended drop zone with a displacement Δr of 3030 m. What is the speed of the bomber when it releases its cargo?

Answer 1

displacement = sqrt(x^2 + y^2)

3030^2 = x^2 + 2430^2

x = 1810 m

In Horizontal,

initial velocity, v0x = v cos36.1

a = 0

x = v0x * t

1810 = (v cos36.1) t

v t = 2240 ........(i)

In vertical,

initial velocity, v0y = - v sin36.1

ay = - 9.8 m/s^2

Applying, y = v0y t + ay t^2 / 2

- 2430 = - (v sin36.1)t - 9.8 t^2 /2

4.9 t^2 + ( v t ) sin36.1 = 2430

putting v t from (i)

4.9 t^2 + 2240 sin36.1 = 2430

4.9 t^2 =1110

t = 15.05 sec

v = 2240/15.05 = 149 m/s .......Ans