Respuesta :

keu
There are 2 cases here:

a. The stone travels to another certain height until its velocity becomes 0.
i.e V² = u² - 2gh (given u = 19.5m/s)

When it reaches maximum height V = 0 
so, h = u²/2g = (19.5)²/(2*9.8) => h= 19.4m

b. Now the body starts to fall freely under gravity, so:
h = 1/2*gt²
We know maximum height reached by the body = height of building + height traveled after throw in (a)

or., t = √(2h/g)
or, t = √2*(19.4+59.4) /9.8 = 4 second
johanrusli

The stone is in the air for 6 seconds.

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Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

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Given:

height of the cliff = h = 59.4 m

speed of the stone = u = 19.5 m/s

Asked:

total time taken = t = ?

Solution:

[tex]h = ut + \frac{1}{2}at^2[/tex]

[tex]-59.4 = 19.5t - \frac{1}{2}(9.8)t^2[/tex]

[tex]-59.4 = 19.5t - 4.9t^2[/tex]

[tex]49t^2 -195t - 594 = 0[/tex]

[tex]( t - 6 ) ( 49t + 99 ) = 0[/tex]

[tex]t - 6 = 0[/tex]

[tex]t = 6 \texttt{ s}[/tex]

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Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Projectile , Motion , Horizontal , Vertical , Release , Point , Ball , Wall

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