Question

A particle is thrown vertically upwards. Its velocity at one-fourth of the maximum height is 20 ms−1.Then, the maximum height attained by it is

Answer 1

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

$\green{\tt{\therefore{Maximum\:height=26.67\:m}}}$

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

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

• According to given question :

$\tt \circ \: Let \: height \: be \: h \\ \\ \tt \circ \: Final \: velocity = 20 \: m/s \: at \: \frac{h}{4} \\ \\ \tt \circ \: Acceleration = - 10 \: {m/s}^{2} \\ \\ \bold{As \: we \: know \: that : } \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {20}^{2} = {u}^{2} + 2 \times ( - 10) \times \frac{h}{4} \\ \\ \tt: \implies 400 = {u}^{2} - 5h \\ \\ \tt: \implies {u}^{2} = 400 + 5h - - - - - (1) \\ \\ \tt \circ \: Final \: velocity = 0 \: m/s \: \: \: (For \: maximum \: height) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {u}^{2} + 2 \times ( - 10) \times h \\ \\ \tt: \implies 0 = {u}^{2} - 20h \\ \\ \tt: \implies {u}^{2} = 20h - - - - - (2) \\ \\ \circ \: \tt{Putting \: value \: of \: {u}^{2} \: in \: (1)} \\ \tt: \implies 20h = 400 + 5h \\ \\ \tt: \implies 15h = 400 \\ \\ \tt: \implies h = \frac{400}{15} \\ \\ \green{\tt: \implies h = 26.67 \: m} \\ \\ \green{\tt \therefore Maximum \: height \: attained \: is \: 26.67 \: m}$

Answer 2

Answer:

Explanation:

Given :-

Velocity at one - fourth height = 20 m/s

To Find :-

Maximum Height Attained by particle.

Formula to be used :-

v² = u² + 2as

Solution :-

Let maximum height be h m.

And the initial velocity is u m.

Final velocity at maximum height , v = 0

Using 3rd equation, v² = u² + 2as

⇒ 0 = u² + 2gh

⇒ u ² = 2gh .... (i)

Final velocity at height   h/4, v = 20 m/s

Using 3rd equation, v² = u² + 2as

⇒ 20² = u² + 2g × h/4

⇒ 20 ² = 2gh + gh/2 (Putting Eq (i))

⇒ h = 80/3

⇒ h = 26.67 m

Hence, the maximum height attained by particle is 26.67 m.