Question

An object thrown vertically upwards reaches a height of 180 m . How long will the object take to come back to the earth ? assume g=10m/s² *

Answer 1

Given :

Maximum height attained by object = 180m

To Find :

Time taken by object to come back to the earth.

Solution :

■ For a body thrown vertically upward, g is taken negative.

First of all we need to find initial velocity of the object$.$

❇ Third equation of kinematics :

➙ v² - u² = 2gH

At highest point, v = 0

➙ 0² - u² = 2(-10)(180)

➙ -u² = -3600

➙ u = √3600

➙ u = 60 m/s

Total time of flight for an object thrown vertically upward at a speed of u is given by

⭆ T = 2u/g

⭆ T = 2(60)/10

⭆ T = 120/10

⭆ T = 12s

∴ Object will come back to the earth after 12s.

Answer 2

Answer:-

$\red{\bigstar}$ Time taken to come back to earth $\large\leadsto\boxed{\rm\green{12 \: seconds}}$

• Given:-

Height of the object when thrown = 180 m

Acceleration = - 10 m/s [Against the gravity]

Final velocity = 0 m/s

• To Find:-

Time taken to come back to earth = ?

• Solution:

• Using 3rd equation of motion:-

$\pink{\bigstar}$ $\large\boxed{\purple{\bf v^2 = u^2 + 2as}}$

Here,

v = Final velocity

u = Initial velocity

a = acceleration [negative]

s = distance travelled

➙ $\sf (0)^2 = (u)^2 + 2 \times (-10) \times 180$

➙ $\sf 0 = u^2 + 2 \times - 1800$

➙ $\sf 0 = u^2 - 3600$

➙ $\sf u^2 = 3600$

➙ $\sf u = \sqrt{3600}$

★ $\bf\red{u = 60 m/s}$

• Using 1st equation of motion:-

➙ $\large\boxed{\purple{\bf v = u + at}}$

➙ $\sf 0 = 60 + (-10)t$

➙ $\sf 0 = 60 - 10 t$

➙ $\sf 10t = 60$

➙ $\sf t = \dfrac{60}{10}$

★ $\bf\red{t = 6 sec.}$

Now,

• Total time taken = Time of ascent + Time of descent

➙ Total time taken = 6 + 6 sec.

➙ Total time taken = $\bf\red{12 sec.}$

Therefore, total time taken to reach earth is 12 sec.