To shake her attacker, Woody flies directly upwards at $21\mathrm{.}0$ km$/$h $($woodpeckers can fly slightly faster than this when flapping their wings, but this particular woodpecker is a little weighted down at the moment$).$ Weasly, unable to hold on$,$ is released from Woodys back at the instant at which they are both $6\mathrm{.}80$ m above the ground. Assume no air resistance for both parts below.

a$)$ Find the position of Weasly $0\mathrm{.}250$ s after his release. Assume that y$=0$ is at the ground.

b$)$ Say that Woody is being chased by another woodpecker, Buzz $($perhaps it's mating season?$).$ If Buzz is directly underneath Woody at a height of $2\mathrm{.}70$ m above the ground when Weasly is released, at what position do Buzz and Weasly collide if Buzz is flying upwards with a constant speed of $18\mathrm{.}0$ km$/$h$?$

Answer rating: 100% (QA)

a To find the position of Weasly 0250 s after his release we first need to calculate the initial vel... View the full answer