Question

consider the vectors a = 27 m pointed at 40 degrees and b = 35 m pointed at 200 degrees both measured ccw from the positive x axis. determine the magnitude of the resultant vector r = a + b, in meters. integer, decimal, or e notation allowed question 4 6 points wanting to measure how fast you can hit a cue ball when you break. so you set up table, 0.93 m above the floor, strike the cue ball, and it flies a horizontal distance of 3.4 m before hitting the floor. with what speed can you hit a cue ball? integer, decimal, or e notation allowed

Explanation:

Step1: Resolve vectors A and B into components

For vector A with magnitude $A = 27$ m and angle $\theta_A=40^{\circ}$:
$A_x = A\cos\theta_A=27\cos40^{\circ}\approx27\times0.766 = 20.682$ m
$A_y = A\sin\theta_A=27\sin40^{\circ}\approx27\times0.643 = 17.361$ m
For vector B with magnitude $B = 35$ m and angle $\theta_B = 200^{\circ}$:
$B_x=B\cos\theta_B=35\cos200^{\circ}\approx35\times(- 0.9397)=-32.8895$ m
$B_y=B\sin\theta_B=35\sin200^{\circ}\approx35\times(-0.342)= - 11.97$ m

Step2: Find the x - and y - components of the resultant vector R

$R_x=A_x + B_x=20.682-32.8895=-12.2075$ m
$R_y=A_y + B_y=17.361-11.97 = 5.391$ m

Step3: Calculate the magnitude of the resultant vector R

$R=\sqrt{R_x^{2}+R_y^{2}}=\sqrt{(-12.2075)^{2}+5.391^{2}}=\sqrt{149.03 + 29.063}=\sqrt{178.093}\approx13.345$ m

for second part:

Step1: Analyze the vertical motion of the cue - ball

The vertical motion of the cue - ball 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 $y=- 0.93$ m (taking downwards as negative).
Using the equation $y = v_{0y}t+\frac{1}{2}at^{2}$, since $v_{0y} = 0$ m/s, we have $y=\frac{1}{2}at^{2}$.
$-0.93=\frac{1}{2}\times(-9.8)t^{2}$
$t^{2}=\frac{2\times0.93}{9.8}=\frac{1.86}{9.8}\approx0.1898$
$t=\sqrt{0.1898}\approx0.436$ s

Step2: Analyze the horizontal motion of the cue - ball

The horizontal motion is a uniform - motion with no acceleration ($a_x = 0$). The horizontal displacement $x = 3.4$ m.
Using the equation $x = v_{0x}t$, we can find the initial horizontal velocity $v_{0x}$ (which is the speed at which the ball is hit).
$v_{0x}=\frac{x}{t}=\frac{3.4}{0.436}\approx7.8$ m/s

Answer:

$13.345$