Aparachutist with a camera descends in free fall at a speed of 10m/s. the parachutist releases the camera at an altitude of 50m. a) how long does it take the camera to reach the ground? b) what is the velocity of the camera before it hits the ground?

a)2.3s b)23m/s

Explanation:

a) According to the equation of motion S = ut+1/2at² where;

S is the distance or the height covered by the object. = 50m

u is the velocity of the object = 10m/s

a= +g(since the body is under a free fall, the acceleration due to gravity will be positive) = 10m/s²

Substituting this value in the formula to get the time we have;

50 = 10t + 1/2(10)t²

50 = 10t + 5t²

5t²+10t-50 = 0

Dividing through by 5 we have;

t²+2t-10 =0

Factorizing and using the general formula we have;

-2±√4+40/2

-2±√44/2

-2+√44/2

t = 2.3seconds

Therefore it takes the camera 2.3seconds to reach the ground.

b) velocity of the camera before it hits the ground can be gotten using the equation of motion

v = u+at

Since a = +g

v = u+gt... 1

The initial velocity of the camera will be zero at the point of drop, therefore its velocity before reaching the ground will be;

v = 0 +10(2.3)

v = 23m/s

i think it would be 5

the components of a vector are the parts of the vectors that are to each other.

b perpendicular