A ball is thrown straight up with an initial speed of 80 m/s. How high does it rise if the acceleration due to gravity is 10 m/s^2?

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The ball is thrown up with a velocity of 80m/s. The acceleration due to the gravitational force of attraction is given as 10 m/s^2 and this acts in the downward direction.

The ball stops moving upwards when its velocity in the upward direction is equal to zero.

We know that...

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The ball is thrown up with a velocity of 80m/s. The acceleration due to the gravitational force of attraction is given as 10 m/s^2 and this acts in the downward direction.

The ball stops moving upwards when its velocity in the upward direction is equal to zero.

We know that Vf = Vi + g*t.

Here Vf = 0, Vi = 80, g = -10m/s^2.

=> 0 = 80 - 10*t

=> t = 80/ 10

=> t = 8 seconds.

So the ball moves upwards for 8 seconds. The distance travelled by the ball in 8 seconds is given by d = Vi*t + (1/2)*g*t^2

=> d = 80*8 - (1/2)*10*8^2

=> d = 640 - 5*64

=> d = 640 - 320

=> d = 320 m

Therefore the ball rises to a height of 320 m.

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