An-object-is-thrown-vertically-upward-from-the-top-of-a-building-20m-high-If-the-object-passes-the-point-it-was-thrown-4-seconds-on-it-way-down-find-the-a-Velocity-at-which-the-object-was-thrown-

Question Number 220581 by Tawa11 last updated on 16/May/25

An object is thrown vertically upward from
the top of a building 20m high. If the object
passes the point it was thrown 4 seconds
on it way down, find the.
(a) Velocity at which the object was thrown.
(b) time taken when the object is 10m above
the level it was thrown.
(c) time taken when the object is 10m below
the level it was thrown.
(d) Velocity with which the object hit the ground.
[Take g = 10m/s²]

Answered by mr W last updated on 17/May/25

$$\left({a}\right) \\ $$$${v}_{\mathrm{0}} ={gt}=\mathrm{10}×\mathrm{2}=\mathrm{20}\:{m}/{s} \\ $$$${or} \\ $$$$\mathrm{0}={v}_{\mathrm{0}} ×\mathrm{4}−\frac{\mathrm{10}×\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{v}_{\mathrm{0}} =\mathrm{20}\:{m}/{s} \\ $$$$\left({b}\right) \\ $$$$\mathrm{10}=\mathrm{20}{t}−\frac{\mathrm{10}{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{t}=\mathrm{2}\mp\sqrt{\mathrm{2}}\approx\mathrm{0}.\mathrm{59}\:{s}\:{and}\:\mathrm{3}.\mathrm{41}\:{s} \\ $$$$\left({c}\right) \\ $$$$−\mathrm{10}=\mathrm{20}{t}−\frac{\mathrm{10}{t}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{t}=\mathrm{2}+\sqrt{\mathrm{6}}\approx\mathrm{4}.\mathrm{45}\:{s} \\ $$$$\left({d}\right) \\ $$$${v}_{\mathrm{2}} =\sqrt{\mathrm{20}^{\mathrm{2}} +\mathrm{2}×\mathrm{10}×\mathrm{20}}\:\approx\mathrm{28}.\mathrm{28}\:{m}/{s} \\ $$

Commented by mr W last updated on 16/May/25

Commented by Tawa11 last updated on 16/May/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{help}\:\mathrm{sir}. \\ $$

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