As he is catching a 2.5 kg package, your friend exerts a downward force of 1.0x10^1 n. if the package was still moving upwards at 6.7 m/s at the start of his catch, how much time will it take to bring him to a stop?

F = ma

A = F / m = 1 x 10^1 + (2.5) (9.3) / 2.5 = 1 x 10^1 + 24.5 /2.5 = 34.5 / 2.5 = 13.8 m/s^2

By using kinematic equation, V = 4 + ak

0 = 6.7 – 13.8 t

13.8t = 6.7

T = 6.7 / 13.8 = 0.4855 sec 

a.

explanation:

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