A plane flying at 147m/s, at 5850m, drops package...?

Question

with 9.8 as accceleration of gravity, where will the plane be when package lands? Air resistence neglected.

1. directly above
2. ahead of package
3.behind package.

and what is the horizontal distance from release point to impact point?


please help. i'm lost >.<

Answer 1

Directly above, for sure, because the package is flying at the same horizontal speed as the plane until it hits the ground.

The horizontal distance is computed by seeing how long it will take for the package to drop, and then multiplying that time by the speed.
Use distance = 1/2 a t^2, so t = sqrt(5850*2/9.8)
and the distance = 147*t

Answer 2

Since the forward velocity of both the plane and the package are the same, the plane will be directly above the package.

Answer 3

initial horizontal v = 147 m/s
initial vertical v= 0 m/s

height = 5850 m

use this formula to find time:
h= 0.5 * g * t^2

t=(h/0.5*g)^0.5


horizontal v is contant
so horizontal distance = v*t
=147 *(h/0.5*g)^0.5

=147 *(5850/0.5*9.81)^0.5
= 5 076.63292m


and plane also covered that much distance in that time.. so 1.

Answer 4

1. directly above

2) time for the package to hit the ground.
t= sqr(2d/g)
t= sqr(2*5850 / 9.8)
t= 34.55s

x=vt
x=(147m/s)(34.55s)
x= 5078.85m