8656211 Assignmen41 Problem 4.40 - Enhanced-with Feedback next a Problem 4.40-En

Question

8656211 Assignmen41 Problem 4.40 - Enhanced-with Feedback next a Problem 4.40-Enhanced - with Feedback Part A On a hot summer day, a young girl swings on a rope above the local swimming hole (Figure 1). When she lets go of the rope her initial velocity is 2.30 m/s at an If she is in flight for 0.612 s, how high above the water was she when she let go of the rope? anale nf 35n°ahnue the hnrizontal Vou may want to Figure 1 011 h1.13 Submit Incorrect: Try Again; 4 attempts remaining Provide FeedbecdkContinue 35.0

Explanation / Answer

here u = 2.30 m/s Q = 35o

so in y direction uy = u sinQ = 2.30 x sin35o

uy = 1.3192 m/s t = 0.612 s

so from newton third eqn of motion in y direction

s = uyt + ayt2 / 2

s = (1.3192 x 0.612 ) - (9.8 x 0.5 x 0.6122 )

s = 0.8073504 - 1.835266

s = -(1.027916) m

we take acceleration positive at downwords so here the negative sign shows the direction of displacement but the hight will remain same . hight dosen't consist direction so

hight H = 1.027916 m or 1.03 m ans

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