The vector c has a magnitude of 24.6m and it is in the negative y direction. Therefore
The vector b is 41.4° up from the x-axis. Therefore
![\vec{b} = b[cos(41.4^{o}) \hat{i} + sin(41.4^{o}) \hat{j} ] =b(0.75\hat{i} + 0.6613 \hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Bb%7D%20%3D%20b%5Bcos%2841.4%5E%7Bo%7D%29%20%5Chat%7Bi%7D%20%2B%20sin%2841.4%5E%7Bo%7D%29%20%5Chat%7Bj%7D%20%5D%20%3Db%280.75%5Chat%7Bi%7D%20%2B%200.6613%20%5Chat%7Bj%7D%29)
The vector a is 27.7° up from the x-axis. Therefore
![\vec{a} = a[cos(22.7^{o})\hat{i} + sin(27.7^{o})\hat{j}] = a(0.8854\hat{i} + 0.4648\hat{j})](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D%20%3D%20a%5Bcos%2822.7%5E%7Bo%7D%29%5Chat%7Bi%7D%20%2B%20sin%2827.7%5E%7Bo%7D%29%5Chat%7Bj%7D%5D%20%3D%20%20a%280.8854%5Chat%7Bi%7D%20%2B%200.4648%5Chat%7Bj%7D%29)
Because
, the sum of the x and y components should be zero. Therefore,
For the x-component,
0.8854a + 0.75b = 0
or
a + 0.847b = 0 (1)
For the y-component,
0.4648a + 0.6613b - 24.6 = 0
or
a + 1.4228b = 52.926 (2)
Subtract (1) from (2).
0.5758b = 52.926
b = 91.917
a = -0.847b = -77.854
Answer:
The magnitude of vector a is -77.85 m
The magnitude of vector b is 91.92 m
The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.
<h3>
What is vertical motion of a projecile?</h3>
The vertical motion of a projectile is affected by gravity and the velocity of vertical motion given by the following formula;
Vy = Vsinθ
<h3>
What is horizontal motion of a projecile?</h3>
The horizontal velocity of a projectile is given by the following formula;
Vx = Vcosθ
<h3>Direction of the motion</h3>
The direction of the motion is calculated as follows;
tanθ = Vy/Vx
Thus, the formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle involve the use of tangent, cosine and sine.
Learn more about vertical motion here: brainly.com/question/24216590
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Tensile strength is the amount of tension a material can hold, at least I hope that’s what it is.
A)acceleration is in the direction of motion
To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.
Here,
v = Final velocity
= Initial velocity
g = Acceleration due to gravity
t = Time
At t = 4s, v = -30m/s (Downward)
Therefore the initial velocity will be
Now the position can be calculated as,
When it has the ground, y=0 and the time is t=4s,
Therefore the cliff was initially to 41.6m from the ground
