An arrow is shot vertically upward and then 1.84s later passes the top of a tree

Question

An arrow is shot vertically upward and then 1.84s later passes the top of a tree 40.9m high. How much longer will the arrow travel upward, and how high will it go?

Explanation / Answer

at tree top, velocity, v = u - g * t1 -------- (1) t1 -> time to reach top of the tree. u -> initial velocity. at top most point, 0 = v - g * t2 -------------- (2) t2 -> time required further to reach the top (required) v*v - u*u = -2g*s ------------------ (3) s-> height of the tree from (1) and (3) solve for "u" u = (g*t1*t1 + 2*s) / (2*t1) = 31.244 m/s hence, v = 13.212 m/s => t2 = 1.348 s for total height reached by the arrow, 0*0 - u*u = -2g*h => h = u*u / 2g = 49.805 m

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