An object is swung in a horizontal circle on a length of string
1 answer:
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Answer:
<em>12 m/s</em>
Explanation:
<u>Projectile Motion</u>
It's also known as 2D motion because the movement takes place in both axis x and y. The x-axis motion is at a constant speed since in absence of friction, no external force stops or accelerates the object. The y-axis motion is at variable speed, which is changed by the acceleration of gravity that makes the object to reach a maximum height and then go back to ground level.
The maximum horizontal distance reached (also called Range) is given by
Knowing that , we solve for Vo
Thus, the initial speed of Mike Powell was 12 m/s
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 vector b is 41.4° up from the x-axis. Therefore
The vector a is 27.7° up from the x-axis. Therefore
Because
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