Question

An object is thrown vertically upwards it has a velocity of 20 m/secwhen it reaches half of its maximum height. how height from the ground does it rise ?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Maximum\:height=40\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies Velocity \: at \: one \: fourth = 20 \: m/s \\ \\ \red{\underline \bold{To \: Find:}}\\ \tt: \implies Maximum \: height \: attained = ?$

• According to given question :

$\tt \circ \: Let \: height \: be \: h \\ \\ \tt \circ \: Final \: velocity = 20 \: m/s \: at \: \frac{h}{2} \\ \\ \tt \circ \: Acceleration = - 10 \: {m/s}^{2} \\ \\ \bold{As \: we \: know \: that : } \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {20}^{2} = {u}^{2} + 2 \times ( - 10) \times \frac{h}{2} \\ \\ \tt: \implies 400 = {u}^{2} - 10h \\ \\ \tt: \implies {u}^{2} = 400 + 10h - - - - - (1) \\ \\ \tt \circ \: Final \: velocity = 0 \: m/s \: \: \: (For \: maximum \: height) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {u}^{2} + 2 \times ( - 10) \times h \\ \\ \tt: \implies 0 = {u}^{2} - 20h \\ \\ \tt: \implies {u}^{2} = 20h - - - - - (2) \\ \\ \circ \: \tt{Putting \: value \: of \: {u}^{2} \: in \: (1)} \\ \tt: \implies 20h = 400 + 10h \\ \\ \tt: \implies 10h = 400 \\ \\ \tt: \implies h = \frac{400}{10} \\ \\ \green{\tt: \implies h = 40\: m} \\ \\ \green{\tt \therefore Maximum \: height \: attained \: is \: 40\: m}$

Answer 2

let it's total height be h

upto h/2

initial velocity=u (say)

final velocity=20

a=g

since the acceleration is uniform,so we can use the law of uniformly acelrated motion

V^2=u^2-2gh/2

400+10h=u^2......I)

when it reach the maximum height

total distance travelled (s)=h

final velocity(V)=0

a=g

using third law of uniformly accelerated motion

0^2=u^2-2×10×h

u^2=20h......ii)

eqauate both the eqaution

20h=400+10h

h=40m

max. heights=40m