Question

An object is thrown vertically upwards and rises to a height of 80 m. Then, the velocity with which the object was thrown upwards is: [ Take, g= 10 m/s2]​

Answer 1

$\\ \\ \large\underline{ \underline{ \sf{ \red{given:} }}}$

• Height attained (s) = 80m

• Final velocity (v) = 0m/sec

• g = -10m/sec² [ as it acts in the opposite direction of motion of ball ]

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}$

• Initial velocity (u)

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

By 3rd Kinematical Equation ,

$\\ \bigstar \boxed{ \bf \: {v}^{2} = {u}^{2} + 2gs}$

Here ,

• v = 0m/sec

• u = ?

• g = -10m/sec²

• s = 80m

Putting values , we get...

$\\ \\ \sf \: {0}^{2} = {u}^{2} + 2( - 10)(80) \\ \\ \sf \: 0 = {u}^{2} + 2( - 800) \\ \\ \sf \: 0 = {u}^{2} - 1600 \\ \\ \sf \: {u}^{2} = 1600 \\ \\ \sf \: u = \sqrt{1600} \\ \\ \boxed{\sf \: \pink{u = 40m {sec}^{ - 1} }}$

Hence , the ball was thrown upward with 40m/sec.

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More to know :-

1st Kinematical equation :-

$\\ \sf \: v = u + at$

2nd Kinematical equatiom :-

$\\ \sf \: s = ut + \frac{1}{2} a {t}^{2}$

Answer 2

Given :

An object is thrown vertically upwards and rises to a height of 80 m.

To find :

The velocity with which the object was thrown upwards. [g= 10 m/s²]​

Solution :

Let the velocity with which object was thrown upwards be "u" m/s

At a height of 80 m, it will be at rest, so final velocity, v = 0 m/s

Gravitational acceleration, g = 10 m/s²

Height reached, h = 80 m

Now as it was thrown upwards, so gravitational acceleration will work against it, so a/c to 3rd equation of motion :

⇒ v² = u² - 2gh

⇒ 0² = u² - 2 × 10 × 80

⇒ 0 = u² - 1600

⇒ u² = 1600

⇒ u = √1600

⇒ u = 40 m/s

∴ Velocity with which it was thrown upwards = 40 m/s.