Question

A ball is thrown upwards from the ground of a tower with a speed of 20 m/s. There is a window in the tower at the height of 15 m from ground when will the ball pass the window during upward motion?(g= 10m/s^2)​

Answer 1

Given:-

A ball is thrown vertically upwards with a speed of 20m/s.

There is a window at height 15m from the ground

To find:-

The time when the ball passes through the window

Explanation:-

Initial velocity,u=20m/s

acceleration,a=-g

Height,S=15m

time,t=?

we know that

$\boxed{\sf S=ut+\dfrac{1}{2}at^2}$

Substituting the values

$= > 15 = 20t - \dfrac{1}{2} g {t}^{2}$

Substituting g=10m/s²

$= > 15 = 20t - \dfrac{1}{2} (10) {t}^{2}$

$= > 15 = 20t - 5 {t}^{2}$

$= > 5 {t}^{2} - 20t + 15 = 0$

$= > {t}^{2} - 4t + 3 = 0$

$= > {t}^{2} - 3t - t + 3 = 0$

$= > t( t - 3) - 1(t - 3) = 0$

$\boxed{ \sf t = 1 \: and \: 3 \: seconds}$

So the body passes through the window at t=1 and t=3 seconds

Answer 2

AnswEr :

• Height of Window = 15 m
• Initial Velocity ( u ) = 20 m/s
• Final Velocity ( v ) = 0 m/s
• Accⁿ due to Gravity ( g) = - 10 m/s²
• Time = ?

• Using 3rd Equation of Motion :

» v² - u² = 2gh

» (0)² - (20)² = 2 × -10 × h

» 0 - 400 = - 20 h

» - 400 = - 20h

• Dividing Both term by - 20

» h = 20 m

⋆ Window is at the height of 15m but ball travelled the distance of 20m. That Means Ball will cross window twice.

• Using 2nd Equation of Motion :

⇒ S = ut + ½ gt²

• Plugging the Values

⇒ 15 = ( 20 × t ) + ( ½ × - 10 × t² )

⇒ 15 = 20t + ( - 5t² )

⇒ 15 = 20t - 5t²

• Dividing Each term by 5

⇒ 3 = 4t - t²

⇒ t² - 4t + 3 = 0

⇒ t² - 3t - t + 3 = 0

⇒ t(t - 3) - 1(t - 3) = 0

⇒ (t - 3)(t - 1) = 0

⇒ t = 3 sec. or, t = 1 sec.

჻ Ball will take 3s to cross window when it will go upward and 1s to Cross window when it will go downwards.