Question

A stone is thrown vertically up from the tower of hight 25m with a speed of 20m/s.what time does it take to reach the ground?(g=10m/s2

Answer 1

It will take 5 seconds to reach the ground.

Solution

As the ston is thrown vertically up from the tower of height 25 m with speed of 20 m/s. gravitational acceleration (g) = 10 m/s²

At first, as the stone is thrown vertically up so gravitational acceleration will work against it.

∴ g = -10 m/s²

While going up :

Here initial speed of stone = 20 m/s

Final speed of stone = 0 m/s

g = -10 m/s²

We will use 1st equation of motion :

● v = u + gt

⇒ 0 = 20 + (-10)t

⇒ 0 = 20 - 10t

⇒ 10t = 20

⇒ t = 20/10

⇒ t = 2 seconds.

∴ Stone will take 2 seconds to go up.

While coming down :

Initial speed = 0 m/s

Height of tower (h) = 25 m

g = 10 m/s²

Maximum height attained by stone :

⇒ u²/2g

⇒ 20²/2 × 10

⇒ 400/20

⇒ 20 m

So, total height = 25 + 20 = 45 m

We will use 2nd equation of motion :

● h = ut + ½ gt²

⇒ 45 = (0 × t) + ½(10) × t²

⇒ 45 = 0 + 5t²

⇒ 45 = 5t²

⇒ t² = 45/5

⇒ t² = 9

⇒ t = √9

⇒ t = 3 s

∵ Time can't be negative.

∴ Time take by stone while coming down = 3 second.

Now total time taken :

⇒ Total time = 2 + 3

⇒ Total time = 5 seconds.

∴ Time taken to reach ground = 5 seconds.

Answer 2

Given :

▪ Height of tower = 25m

▪ Initial speed = 20mps

To Find :

▪ Time taken by ball to reach the ground.

Concept :

✏ Since, acceleration due to gravity has said to be constant, we can easily apply equation of kinematics to solve this type of question.

✏ For a body falling freely under the action of gravity, g is taken positive.

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

✴ For a body thrown vertically up with initial speed u,

✒ Time to ascent :

$\underline{\boxed{\bf{\red{T=\dfrac{u}{g}}}}}$

✒ Maximum height reached :

$\underline{\boxed{\bf{\blue{H=\dfrac{u^2}{2g}}}}}$

✴ For a freely falling body :

$\underline{\boxed{\bf{\green{H=\dfrac{1}{2}gt^2}}}}$

Calculation :

▶ Time taken by ball to cover distance AB :

→ t = u/g

→ t = 20/10

→ t = 2s

▶ Maximum height attained by ball :

→ h = u$^2$/2g

→ h = (20×20)/(2×10)

→ h = 20m

[Height from ground = 25+20 = 45m]

▶ Time taken by ball to cover BC :

→ h = (1/2)gt'$^2$

→ t' = √(2×45/10)

→ t' = √(90/10)

→ t' = √9

→ t' = 3s

⏭ Total time taken by ball to reach at ground T = t + t' = 2 + 3 = 5s

➡ Total time = 5s