Answer to Question #260198 in Physics for grey tyrant

Question #260198

A parkour artist jumps from one rooftop to another. He jumps off the first rooftop with a velocity of 5m/s, angled at 30° to the horizontal. He stays airborne for 0.70 seconds before landing on the second rooftop.





c. What vertical displacement (∆𝑦⃗) did the parkour artist travel when jumping between the


1
Expert's answer
2021-11-05T18:19:15-0400

The time during which the body reached its maximum height


t=vsin30°/g=5sin30°/9.81=0.255 (s)t=v\sin30°/g=5\cdot\sin30°/9.81=0.255\ (s)


t1=0.70.255=0.445 (s)t_1=0.7-0.255=0.445\ (s)


s1=gt2/2=9.80.2552/2=0.32 (m)s_1=gt^2/2=9.8\cdot0.255^2/2=0.32\ (m)


s2=gt12/2=9.80.4452/2=0.97 (m)s_2=gt_1^2/2=9.8\cdot0.445^2/2=0.97\ (m)


Δy=0.32j0.97j=0.65j\Delta \vec y=0.32 \vec j-0.97 \vec j=-0.65\vec j




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