Question

The length of is 3.05 m and the angle θ = 25.0°. Express the resultant A + B in unit-vector notation.

Answer 1

Projectile Motion?? Not enough information for me to understand what you are asking

Answer 2

A + B= 5.102588969
= 5.1

Direction
Angle in the triangle=\[32^{0} 11^{1} 54.02^{11}\]
Angle of elevation from horizontal = \[57^{0} 48^{1} 5.98^{11}\]


Answer= 5.1 \[57.8^{0}\] ???

Answer 3

first consider the x and y components of A and B
$$A_x=|\vec A|\cos \theta,\quad A_y=|\vec
A|\sin \theta$$$$B_x=|\vec B|\cos \alpha,\quad B_y=|\vec B|\sin \alpha$$ In here, since B is horizontal you can take straight forward that , $$B_y=|\vec B|, B_x=0$$Now you can find the resultant vector by adding each component
If, $$\vec R=\vec A+\vec B\\
R_x=A_x+B_x, \quad R_y = A_y+B_y$$
The resultant will be $$|\vec R|=\sqrt{R_x^2+R_y^2}$$ IF the angle is beta,
$$\beta=\tan^{-1}\left(\frac{R_y}{R_x}\right)$$

$$R_x=3\cos(25)=2.71\\ R_y=3.05+3\sin(25)=4.31\\ |\vec R|=5.1\\\alpha =57.8^\circ$$

$$\vec R=5.1 \mathbf{n}$$ where $$\mathbf{n}=(\cos \alpha \;\mathbf i, \sin \alpha\; \mathbf j)$$

Answer 4

Thanks so much!!