A package is dropped from a helicopter moving upward at 15m/s. If the helicopter was 1000m above the ground when the package was released, how long does it take before the package strikes the ground

yi = Initial height of the helicopter

yf = final height of the helicopter

vyi = component of the initial vertical velocity of the helicopter

g = gravity constant (9.8m/s^2)

yf = yi + vyideltat - 1/2gt^2

0m = 1000m + (15m/2)deltat - 1/2(9.8m/s^2)t^2

-1000m = (15m/s)t - (-4.9m/s^2)t^2

Use the quadratic formula

4.8t^2 - 15t - 1000 = 0

t1 = 15.75s and t2 = -12.65

t2 is rejected, time can't be negative

Thus, it takes 15.75s before the package strikes the ground.

because growing process had stoped after 4-5 year

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west