Question

consider the vectors a = 15 m pointed 140 degrees and b = 12 m pointed 30 degrees both measured ccw from the positive x axis. determine the magnitude of the resultant vector r = a + b, in meters. 15.68 integer, decimal, or e notation allowed question 4 you launch a projectile horizontally from a height of 0.85 m above the ground. the projectile travels a horizontal distance 1.8 m before landing on the ground (at a height of 0 m). with what speed (m/s) did you launch the projectile? 4.32 integer, decimal, or e notation allowed

Explanation:

Step1: Resolve vector A into components

$A_x = 15\cos(140^{\circ})\approx - 11.49$ m, $A_y=15\sin(140^{\circ})\approx 9.64$ m

Step2: Resolve vector B into components

$B_x = 12\cos(30^{\circ})\approx 10.39$ m, $B_y = 12\sin(30^{\circ}) = 6$ m

Step3: Find x - component of resultant vector

$R_x=A_x + B_x=-11.49+10.39=-1.1$ m

Step4: Find y - component of resultant vector

$R_y=A_y + B_y=9.64 + 6=15.64$ m

Step5: Calculate magnitude of resultant vector

$R=\sqrt{R_x^{2}+R_y^{2}}=\sqrt{(-1.1)^{2}+15.64^{2}}\approx 15.68$ m

for second question:

Step1: Analyze vertical motion

The vertical - motion of the projectile is a free - fall motion. The initial vertical velocity $v_{0y} = 0$ m/s, the acceleration due to gravity $g = 9.8$ m/s², and the vertical displacement $y - y_0=- 0.85$ m. Using the equation $y - y_0=v_{0y}t+\frac{1}{2}at^{2}$, we have $-0.85 = 0\times t-\frac{1}{2}\times9.8t^{2}$. Solving for $t$:
$4.9t^{2}=0.85$, $t=\sqrt{\frac{0.85}{4.9}}\approx0.42$ s

Step2: Analyze horizontal motion

The horizontal displacement $x - x_0 = 1.8$ m. In horizontal motion (no acceleration, $a_x = 0$), the equation is $x - x_0=v_{0x}t$. We know $t\approx0.42$ s and $x - x_0 = 1.8$ m. Solving for $v_{0x}$:
$v_{0x}=\frac{x - x_0}{t}=\frac{1.8}{0.42}\approx4.32$ m/s

Answer:

15.68