Question

a ball is thrown vertically upward from the top of a tower with a velocity of 20m/s . the ball strikes the earth after 5 s of its throwing . what is the height of the tower? what will be the velocity of the ball while striking the Earth?​

Answer 1

Answer

Height = 25 m

Velocity = 30 m/s

Given

A ball is thrown vertically upward from the top of a tower with a velocity of 20 m/s and the ball strikes the earth after 5 s of its throwing

To Find

Height of the tower

Velocity of the ball while striking the earth

Solution

Find attachment

Initial velocity , u = 20 m/s

Time , t = 5 s

For AB :

Initial velocity , u = 20 m/s

Final velocity , v = 0 m/s

Since , finally moves to rest

Acceleration due to gravity , a = -10 m/s²

Since , thrown against the gravity

Time , t' = ? s

Distance , s' = ? m

Apply 1st equation of motion ,

⇒ v = u + at

⇒ (0) = (20) + (-10)t

⇒ 10t = 20

⇒ t = 2 s

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (0)² - (20)² = 2(-10)s'

⇒ -400 = -20s'

⇒ s' = 20 m

For BD :

Initial velocity , u = 0 m/s

Since , starts from rest

Final velocity , v = ? m/s

Time , t" = t - t' = 5 - 2 = 3 s

Acceleration due to gravity , a = 10 m/s²

Distance , s" = ? m

Apply 1st equation of motion ,

⇒ v = u + at

⇒ v = (0) + (10)(3)

⇒ v = 30 m/s

This is the velocity of the ball while striking the ground .

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (30)² - (0)² = 2(10)s"

⇒ 900 = 20s"

⇒ s" = 45 m

Height of the tower ,

H = s" - s'

⇒ H = 45 - 20

⇒ H = 25 m

Answer 2

Answer:-

Given:-

• Total time the ball took to reach at the ground = 5 seconds [t]
• Initial velocity of the ball = 20 m/s [u]

To Find:

1. Height of the tower [h] &
2. Final velocity of the ball [v].

(Acceleration due to gravity = ± 10 m/s²)

• Time taken to go from I to H:-

Final velocity becomes (v) = 0 m/s

Acceleration becomes (a) = - 10 m/s² [Upward]

Initial velocity (u) = 20 m/s

We know,

v = u + at

or, t = (v - u)/a

In this case,

t = - u/a (Since v = 0 m/s)

Or,

t = [- 20 m/s]/[- 10 m/s²]

→ t = 2 seconds

∴ It took 2 seconds to go from I to H.

Since the total time taken was = 5 seconds,

∴ It took (5 - 2) seconds or, 3 seconds to go from I to H.

• Finding final velocity of the ball:- (H to V)

Initial velocity (u) = 0 m/s

Acceleration due to gravity (a) = + 10 m/s² [Downward]

Time taken (t) = 3 seconds

We know,

v = u + at

In this case,

v = at (Since u = 0 m/s)

→ v = (10 m/s²)(3 s)

→ v = 30 m/s

∴ The velocity the ball touched the Earth was of 30 m/s.

• Finding Height of the Tower:-

HV:-

We know,

h = ut + ½gt²

or, h = ½at² (In this case)

→ h = ½(10 m/s²)(3 s)²

→ h = (5 m/s²)(9 s²)

→ h = 45 m

IH:-

2gh = v² - u²

→ h = - u²/2g

→ h = - (20 m/s)²/[2 × - 10 m/s²]

→ h = (400 m²/s²)/(20 m/s²)

→ h = 20 m

Here,

Actual height of the tower would be = HV - IH

→ 45 m - 20 m

→ 25 m

∴ The height of the tower was of 25 metres.