6.0 seconds after launch, the space shuttle is 529.2 m above the ground.

see explanation below to answers

Explanation:

According to what you say in the comments, you need several stuff to get calculated.

First, we know the distance that the shuttle has after 6 seconds.

We can assume that the initial speed of the shuttle is zero because at time 0, the shuttle is still on ground.

The equation to get the distance with only time and acceleration is:

y = at^2 / 2

Solving for a:

a = 2y/t^2

replacing the values we have:

a = 2 * 529.2 / 6^2

a = 29.4 m/s^2

For the speed of the shuttle after 3 seconds:

Vf = at

Vf = 29.4 * 3 = 88.2 m/s

the speed after 6 seconds:

Vf = 29.4 * 6 = 176.4 m/s

The average speed is:

Vavg = 88.2 + 176.4 / 2 = 132.3 m/s

Finally to get the high of the shuttle in 3 seconds:

y = at^2 / 2

y = 29.4 * 3^2 / 2

y = 132.3 m

d.7.5cm is your answer

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