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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

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.