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Question Number 209432 by Tawa11 last updated on 10/Jul/24
A body is projected  vertically upwards  with    a speed  of 20m/s. Find the  time in seconds    when  the  body is 15m above it point of projection.      g= 10m/s²
A body is projected vertically upwards with

a speed of 20m/s. Find the time in seconds

when the body is 15m above it point of projection.

g= 10m/s²

Answered by A5T last updated on 10/Jul/24
s=ut−((gt^2 )/2)⇒15=20t−5t^2 ⇒t^2 −4t+3=0  ⇒t=3 or 1
$${s}={ut}−\frac{{gt}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow\mathrm{15}=\mathrm{20}{t}−\mathrm{5}{t}^{\mathrm{2}} \Rightarrow{t}^{\mathrm{2}} −\mathrm{4}{t}+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{3}\:{or}\:\mathrm{1} \\ $$
Commented by Tawa11 last updated on 10/Jul/24
Sir, at maximum height,  t = 2 sec.  So, t = 1 sec is the answer?
$$\mathrm{Sir},\:\mathrm{at}\:\mathrm{maximum}\:\mathrm{height},\:\:\mathrm{t}\:=\:\mathrm{2}\:\mathrm{sec}. \\ $$$$\mathrm{So},\:\mathrm{t}\:=\:\mathrm{1}\:\mathrm{sec}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}? \\ $$
Commented by A5T last updated on 10/Jul/24
At t=3; the body is also 15m above the point of  projection.
$${At}\:{t}=\mathrm{3};\:{the}\:{body}\:{is}\:{also}\:\mathrm{15}{m}\:{above}\:{the}\:{point}\:{of} \\ $$$${projection}. \\ $$
Commented by Tawa11 last updated on 10/Jul/24
Sir, but why at maximum height  t = 2 sec.  So, the 3 sec is bigger than the time at maximum  height, can I say t = 1 sec. only
$$\mathrm{Sir},\:\mathrm{but}\:\mathrm{why}\:\mathrm{at}\:\mathrm{maximum}\:\mathrm{height} \\ $$$$\mathrm{t}\:=\:\mathrm{2}\:\mathrm{sec}. \\ $$$$\mathrm{So},\:\mathrm{the}\:\mathrm{3}\:\mathrm{sec}\:\mathrm{is}\:\mathrm{bigger}\:\mathrm{than}\:\mathrm{the}\:\mathrm{time}\:\mathrm{at}\:\mathrm{maximum} \\ $$$$\mathrm{height},\:\mathrm{can}\:\mathrm{I}\:\mathrm{say}\:\mathrm{t}\:=\:\mathrm{1}\:\mathrm{sec}.\:\mathrm{only} \\ $$
Commented by A5T last updated on 10/Jul/24
If we are considering the whole motion, after  t=2 the body will start to drop down from 20m.
$${If}\:{we}\:{are}\:{considering}\:{the}\:{whole}\:{motion},\:{after} \\ $$$${t}=\mathrm{2}\:{the}\:{body}\:{will}\:{start}\:{to}\:{drop}\:{down}\:{from}\:\mathrm{20}{m}. \\ $$
Commented by Tawa11 last updated on 10/Jul/24
Thanks sir, I understand
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{understand} \\ $$
Commented by Tawa11 last updated on 10/Jul/24
sir help me check Q209430
$$\mathrm{sir}\:\mathrm{help}\:\mathrm{me}\:\mathrm{check}\:\mathrm{Q209430} \\ $$

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