Question

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

Answer 1

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:

$d_x = 12 km$ east

$d_y = 16 km$ south

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

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km$

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

$S=\frac{1}{2}gt^2$

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

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s$

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

$v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s$

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

$t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s$

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

$d=vt=(16 m/s)(2 s)=32 m$

So, the closest answer is B.