Question

A ball is thrown upwards from the ground at a tower with a speed of 20 m/s. There is a window in the tower at the height of 15 m from the ground. How many times and when will the ball pass the window? [Take g= 10 m/s]​

Answer 1

Answer:

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

t=1 and 3 sec

Explanation:-

Initial velocity,u=20m/s

acceleration,a=-g

Height,S=15m

time,t=?

we know that

Substituting the values

=  > 15 = 20t -1/2gt ²

Substituting g=10m/s ²

=  > 15 = 20t - 1/2[10t²]

=  > 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

 t=1 and 3 sec

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

Explanation:

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Answer 2

$\huge\mathfrak{Bonjour!!}$

$\huge\mathbb\purple{Solution:-}$

⛄ Given,

★ Initial velocity, u= 20 m/s

★ Maximum height the ball will reach, h= ?

Now,

Using equation, v²= u² + 2gh [Since at the maximum height, v= 0]

=) h= -u²/2g = -(20)²/2(-10)

=) h= 400/20 = 20m.

☞ This means, the ball will reach the height of 20 m and come back. Also, it will pass the window 2 times.

Now,

to calculate the time the ball will take to reach the height of 15 m,

Using equation, h= ut - 1/2 gt²

=) 15 = 20t - 1/2 (10)t²

=) 5t² - 20t + 15 = 0

=) t² - 4t + 3 = 0

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

Therefore,

☞ t= 1,3.

Thus, the ball will pass the window at 1 second and 3 seconds respectively.

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