1-a car moves toward east 12km is represented as a and it turns towards south 16km is represented as b. what is the resultant vector of a-b
a-20 km south east
b-25 km south east
c-20km north east
d-25 km north east

2-a stone is thrown horizontally from the top of a tower of height 78.4 m. it reaches 40 m away from the foot of the tower. what is its initial velocity of projection?
h=1/2gt^2
x=vx^2
a=2.5 m/s
b=5 m/s
c=7.5 m/s
d=10 m/s

a storm is projected horizontally with a velocity of 16 m/s from the bridge at a height of 20 m from the river. how far away (x) parentheses from the bridge is the stone when it reaches the river
h= 1/2gt^2
x=vx^2
a= 36 m
b=32.32 m
c=4m
d=24.23 m

1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

east

south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

From which we find the total time of the fall, t:

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is

So, the closest answer is B.