Question

An object projected upwards acquires a velocity of 9.8 m/s when it reaches half of the

maximum height. Find the maximum height reached by the object​

Answer 1

Answer:

Assuming "projected upwards" means "projected VERTICALLY upwards". Then when falling from its max height (h) the object will achieve the same downward velocity = 9.8 m/s as it had when vertically rising at h/2.

Explanation:

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Answer 2

MAX HEIGHT BY OBJECT IS 3.2 METRES.

Given:

• At half of max height, velocity of the object is 9.8 m/s.

To find:

• Max height reached by object?

Calculation:

• Let's assume that initial velocity is u and max height is h.

So, as per equation of kinematics:

${v}^{2} = {u}^{2} + 2( - g)h$

• Here , at max height h , final velocity will be zero.

$\implies {0}^{2} = {u}^{2} - 2gh$

$\implies h = \dfrac{ {u}^{2} }{2g} \: \: \: \: ...(1)$

Now, at half of max height :

${v}^{2} = {u}^{2} + 2( - g) \dfrac{h}{2}$

$\implies {9.8}^{2} = {u}^{2} - gh$

$\implies h = \dfrac{{9.8}^{2} - {u}^{2}}{g} \: \: \: \: \: \: \: ...(2)$

Now, equating (1) and (2) :

$\dfrac{ {u}^{2} }{2g} = \dfrac{ {9.8}^{2} - {u}^{2} }{g}$

$\implies \dfrac{ {u}^{2} }{2} = {9.8}^{2} - {u}^{2}$

$\implies 3{u}^{2} = 192.08$

$\implies {u}^{2} \approx 64$

$\implies u \approx 8 \: m {s}^{ - 1}$

Now, putting value of 'u':

$\implies h = \dfrac{ {u}^{2} }{2g}$

$\implies h = \dfrac{ {(8)}^{2} }{2(10)}$

$\implies h = 3.2 \: m$

So, maximum height achieved by the object is 3.2 metres.