Question

An airplane with a speed of 53.6 m/s is climbing upward at an angle.....?

An airplane with a speed of 53.6 m/s is climbing upward at an angle of 40° with respect to the horizontal. When the plane's altitude is 820 m, the pilot releases a package.

(a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.

____ m

(b) Relative to the ground determine the angle of the velocity vector of the package just before impact.

____ ° clockwise from the positive x axis

Answer 1

See the diagram,the package will follow the pathway shown below,

Considering the airplane is constantly rising at an angle of `40`degree w.r.t horizontal at a speed `53.6 m/s`

So,at point A, `820 m` above the ground the pilot releases the package,it will follow a projectile motion upto B then fall down due to gravity,but at the same time move forwards(GC),due to the horizontal component of the initial velocity.

So,total distance travelled by the package,along ground is FC =AB+GC (from the diagram)

Now, AB= range of projectile motion = `(u^2 sin 2 theta)/g`

Given `u=53.6 m/s , theta=40`

So, AB=`282.93 m`

Now, vertical height=BG=`820m`,if it takes `t s` to fall by that distance,then we can apply,

`s=ut+1/2 at^2`

Here, `s=820,u=53.6 sin 40,a=g` (as when the package will return to point B,it will have the same value of velocity downwards,that it had upwards at point A...talking about vertical component of velocity only,horizontal one is fixed always)

So,`t = 9.8 s`

So, GC = `u cos theta *t` ( as this length of path was travelled with horizontal component of initial velocity)

So, GC = `402.38m`

So,total distance travelled along ground is (AB+GC) `=(282.93+402.38)m =685.31 m`

Now, if during hitting the ground its velocity was `v m/s` vertically,then we can write, `v=u sin 40+g t`

So, `v = 132.45 m/s`

So,net velocity while striking the ground is `sqrt(132.45^2+(u cos 40)^2) m/s` i.e `138.668 m/s` directed at an angle `tan^-1(u cos 40/132.45)` i.e `17.22` degrees w.r.t vertical direction

So, w.r.t ground clockwise the angle of net velocity is `(90-17.22)` degrees or,`72.78` degrees