An airplane flying horizontally at a constant speed of 350 km/h over level ground releases a bundle of food supplies. ignore the effect of the air on the bundle. what is the bundle’s initial vertical component of velocity? enter your answer in accordance to the question statement 0 km/hr what is the bundle’s initial horizontal component of velocity? enter your answer in accordance to the question statement 350 km/hr what is its horizontal component of velocity just before hitting the ground? enter your answer in accordance to the question statement km/hr if the airplane’s speed were, instead, 450 km/h, would the time of fall be longer, shorter, or the same?

Explanation:

Bundle's initial  vertical velocity is zero because the plane is  moving horizontally and its vertical component is zero. 0km/hr

Bundle's initial and uniform horizontal velocity is 350 km/hr .

The Bundle will maintain its horizontal velocity all along its journey until it hits the ground. It is so because it faces no force opposite to it. The only force that is acting on it is vertical gravitational  force which increases its velocity vertically and not horizontally. So the horizontal component of its velocity just before hitting the ground is 350 km/hr.

If the horizontal velocity of plane is increased to 450 km/hr . there will be no effect on time of fall of the bundle  because horizontal velocity has nothing to do with vertical displacement. vertical displacement is due to gravitational acceleration g. So time of fall will be the same.

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j