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hey is this a calc/derivative problem?
or are you supposed to be solving it in a different way
you can do this with the first derivative
you want to know when the rate of change of the height is zero, so you take the first derivative
1. h = -4.9t^2 + 56t 2. dh/dt = -4.9*2*t + 56
3. 56 -9.8t
there will be a maximum when that rate is zero
so 56 = 9.8t
t - 56/9.8
errr t = 56/9.8
= 560/98 = 280/49 = 40/7
so at 40/7 seconds, it will reach its highest point
so 40/7 seconds is the correct answer for (B). if you put that back into original height equation, you will have the answer to A.
the answer to C is just 3 meters plus the answer to A
since the rocket still peaks after the same amount of time - just three meters higher
let me know if you were supposed to do this in a different way (eg. graphing)
can you do it graphing? That would be a great help to have both methods.