Question

$\huge\mathfrak{QuEStiON}$ :
An object is thrown vertically upwards in such a way that it has a speed of 19.6 m/s when it reaches half of its maximum height. Find the
1) maximum height
2) velocity after 1 second it is thrown
3) it's acceleration when the object reaches maximum height.
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Answer 1

Maximum height

Let maximum height be h. At maximum height, final velocity = 0

v'² - v² = -2gh' (-ve term as g acts downward)

0 - v² = -2gh'

v² = 2gh'

h' = h/2

$v = \sqrt{2gh'}$

$19.6 = \sqrt{9.8h}$

19.6² = 9.8h

h = 39.2m

Velocity after 1s of throwing:

Initial speed, u = √(2gh)

h = maximum height

u = 19.6√2

Velocity after 1s,

v'' - u = -gt (-ve term as g acts downward)

v'' = 9.8 (2√2 - 1)

Acceleration:

Acceleration at all points is g = 9.8

Answer 2

hi ....

I'm from Kerala

Maximum height

Let maximum height be h. At maximum height, final velocity = 0

v'² - v² = -2gh' (-ve term as g acts downward)

0 - v² = -2gh'

v² = 2gh'

h' = h/2

v = \sqrt{2gh'}v=

2gh

′

19.6 = \sqrt{9.8h}19.6=

9.8h

19.6² = 9.8h

h = 39.2m

Velocity after 1s of throwing:

Initial speed, u = √(2gh)

h = maximum height

u = 19.6√2

Velocity after 1s,

v'' - u = -gt (-ve term as g acts downward)

v'' = 9.8 (2√2 - 1)

Acceleration:

Acceleration at all points is g = 9.8