A toy rocket, launched from the ground, rises vertically with an acceleration of 13 m/s2 for 7.6 s until its m

This is actually simple. Let me explain for your better comprehension about this problem.

The basic idea behind is that 13 m/s^2 shows that
in 1 second, the toy's speed changes by 13 m/s
so, in 7.6 seconds, the speed changed by 98.8 m/s (Average speed)
And to find the final height it reaches, use formula

Average speed = Total distance/ Total time
You get
Total distance = speed x time
= 98.8 x 7.6
= 750.88 m
OR

You can use the formula:

a = (v - u)/t

where a --> acceleration of toy = 13 m/s^2
v --> final velocity of toy
u --> initial velocity of toy
t --> time taken for change in velocity = 7.6 s

Yet. here v - u would be the same as a change in velocity, that is the total increase in speed of the toy as it rises in the air and reaches its final height.

So, v - u can be written as 'v' itself for speed.

Replace the values:

13 = v/7.6
v = 13 x 7.6
= 98.8 m/s

Now it's easy to calculate the distance moved.

From speed = distance/time
distance = speed x time
= 98.8 x 7.6
= 750.88 m

To 3 significant figures, 751 m up in the air.
Hope this helps :)
For more explanation and practice, go to:

http://www.physicsforums.com/showthread.… or
http://answers.google.com/answers/thread…

I think there is probably more to this question... Do you want to know the height after 7.6 seconds?