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tatuchka [14]
3 years ago
5

A ball is thrown up at 30m/s from the ground. What is it s maximum height? How long did it take?

Physics
2 answers:
umka21 [38]3 years ago
8 0

Answer:

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

<em>The maximum height of the ball is 45.92 meters</em>

<em></em>

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

<em>The time to reach the maximum height is 3.06 seconds </em>

Rzqust [24]3 years ago
3 0

<u>Answer: </u>

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

<u>Explanation</u>:

Given:

Initial velocityu=30 \mathrm{m} / \mathrm{s}

To find :

maximum height =?

time taken=?

Solution:

<em>Step 1:Finding the time taken to reach the highest point</em>:

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

Using the formula,

v=u+a t

Where a is acceleration due to gravity.Its value is-9.8 m / s^{2}

Substituting the values:  

0=30+(-9.8) t

-30=(-9.8) t

t=\frac{30}{9.8}

t=3.06 \mathrm{seconds}

<em>Step 2: finding the highest point </em>

Using the formula

v^{2}-u^{2}=2 a s

where

s is the maximum highest.

Substituting values

0^{2}-(30)^{2}=2(-9.8) s

0-900=(-19.6) s

-900=(-19.6) s

s=\frac{900}{19.6}

s=45.9 \text { meters }

Result:

Thus the maximum height reached is 45.9 meters in 3.06 seconds

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