Aball is projected upward at time t = 0.0 s from a point on a roof 90 m above the ground. the ball rises, then falls and strikes the ground. the initial velocity of the ball is 80.5 m/s. the velocity of the ball when it is 89 m above the ground is about : im making this short and simple .

The velocity of the ball when it is 89 meters above the ground is about: E. - 81 m/s

Given the following data:

To find velocity of the ball when it is 89 m above the ground, we would use the third equation of motion;

Mathematically, the third equation of motion is given by the formula:

Velocity, V = 80.62 ≈ 81 m/s

But since the ball is falling, the velocity would be negative.

Velocity, V =  - 81 m/s

Therefore, the velocity of the ball when it is 89 meters above the ground is about - 81 m/s.

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Option D.

Step-by-step explanation:

A ball is projected upward with an initial velocity = 80.5 m per second.

Height of the roof from the ground is 90 m.

Equation that represents the relation between velocity and height when thrown upwards is

Here final velocity v = 0, initial velocity u = 80.5 meter per second and g = 9.8 m per second²

0 = (80.5)² - 2(9.8)h

19.6h = 6480.25

h = meters

Total height of the ball from the ground = 90 + 330.625 = 420.625 m

Now we have to calculate the velocity of the ball at the height = 89 m

Distance covered by the ball falling freely form the point above 420.625

Distance covered by the ball above 89 m = 420.625 - 89 = 331.625 m

We plug in the values in the formula

v² = u² + 2gh

v² = 0 + 2(9.8)(331.625)

v = √6499.85 = 80.62 m

  ≈ 81 m

And the sign notation will be negative

Therefore option D. (-81 m) is the answer.

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step-by-step explanation:

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