Question

the ball is thrown vertically upwards with a velocity of 39 m/s calculate the maximum height to which it rises brainliest​

Answer 1

Answer :

• The maximum hieght reached by the ball with the velocity of 38 m/s is

Explanation :

Given :

• Initial velocity of the ball, u = 39 m/s

• Acceleration due to gravity acting on the ball, g = 10 m/s² (Here, g = -10 m/s²)

[Acceleration due to gravity will be taken as negative, since in this case , the object is falling against the gravity]

• Final velocity of the ball, v = 0

[At the maximum hieght, final velocity of a object is 0 .i.e, v = 0]

To find :

• Maximum hieght reached by the ball, h = ?

Knowlwdge required :

Third equation of motion :

[Under gravity]

⠀⠀⠀⠀⠀⠀⠀⠀⠀v² = u² - 2gh

Where,

• v = Final velocity of the object
• u = Initial velocity of the object
• g = Acceleration due to gravity
• h = Height

Solution :

By using the third equation of motion and substituting the values in it, we get :

⠀⠀=> v² = u² - 2gh

⠀⠀=> 0² = 39² - 2 × 10 × h

⠀⠀=> 0 = 39² - 20h

⠀⠀=> -39² = - 20h

⠀⠀=> 1521 = 20h

⠀⠀=> 1521/20 = h

⠀⠀=> 76.05 = h

⠀⠀⠀⠀⠀⠀⠀∴ h = 76.05 m

Therefore,

• Maximum hieght reached by the ball, h = 76.05 m

Answer 2

Given :-

• Initial velocity = 39 m/s
• Final velcoity = 0 m/s ( because the object is reached at maximum height )
• Acceleration = -9.8 m/s²

To find :-

• Maximum height

Solution :-

We are given Initial velocity, final velocity and acceleration due to gravity, By substituting the value in 3rd equation of motion, we can find the value of height.

• u = 39 m/s
• v = 0
• a = -9.8 m/s²

➩ $\rm v^2 = u^2 + 2as$

➩ $\rm 0 = 39^2 - 2 \times 9.8 \times s$

➩ $\rm 1521 = 19.6s$

➩ $\rm s = \dfrac{1521}{19.6}$

➩ $\rm s = 76.8$

Maximum height = 76.8 m