In a circus trapeze act, two acrobats actually fly through the air and grab on t

Question

In a circus trapeze act, two acrobats actually fly through the air and grab on to each other, then together grab a swinging bar. One acrobat, with a mass of 48 kg, is moving at 3.4 m/s at an angle of 14° above the horizontal and the other, with a mass of 51 kg, is approaching her with a speed of 3.2 m/s at an angle of 18° above the horizontal. What is the direction and speed of the acrobats right after they grab on to each other? Let the positive xaxis be in the horizontal direction and assume the first acrobat has positive velocity components in the positive x and ydirections.

The final velocity of the acrobats? (m/s)

?° above the xaxis.

please show all steps

Explanation / Answer

Here,

let the final velocity is v

v * (48 + 51) = 48 * 3.4 * (cos(14) i + j * sin(14)) + 51 * 3.2 * (-cos(18) i + j * sin(18))

solving for v

v = 0.032 i + 0.91 j m/s

speed = sqrt(0.032^2 + 0.91^2) = 0.91 m/s

angle = arctan(0.91/.032) = 88 degree

the speed of the acrobats is 0.91 m/s at 88 degree above horizontal

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects