Aball is thrown up at 30m/s from the ground. what is it s maximum height?
how long did it take?

Maximum height reached by the ball thrown up at  30m/s is   3.06 seconds  and the height reached is  45.9 meters

Explanation:

Given:

Initial velocity

To find :

maximum height =?

time taken=?

Solution:

Step 1:Finding the time taken to reach the highest point:

The velocity of the ball at its highest ,final velocity v=0

Using the formula,

Where a is acceleration due to gravity.Its value is

Substituting the values:  

Step 2: finding the highest point

Using the formula

where

s is the maximum highest.

Substituting values

Result:

Thus the maximum height reached is 45.9 meters in 3.06 seconds

The maximum height of the ball is 45.92 meters

The time to reach the maximum height is 3.06 seconds

Explanation:

A ball is thrown up at 30 m/s from the ground

We need to find its maximum height and how long it took

At maximum height speed equal zero

The acceleration of gravity is -9.8 m/s²

Lets find a rule contains distance, initial speed, final speed and

acceleration

→ v² = u² + 2 g h

where v is the final speed , u is the initial speed, g is the acceleration

of gravity and h is the height

→ v = 0 m/s , u = 30 m/s , g = -9.8 m/s²

Substitute these values in the rule above

→ (0)² = (30)² + 2(-9.8)(h)

→ 0 = 900 - 19.6 h

Add 19.6 h to both sides

→ 19.6 h = 900

Divide both sides by 19.6

→ h = 45.92 m

The maximum height of the ball is 45.92 meters

We need to find the time of the maximum height, then lets use the rule

→ v = u + g t

→ v = 0 m/s , u = 30 m/s , g = -9.8 m/s²

Substitute these values in the rule above

→ 0 = 30 - 9.8 t

Add 9.8 t to both sides

→ 9.8 t = 30

Divide both sides by 9.8

→ t = 3.06 seconds

The time to reach the maximum height is 3.06 seconds

found in the nucleus

has mass of one amu

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