Answer:
73.67 m
Explanation:
If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:
,
Where
its our initial height,
our initial speed, a the acceleration and t the time that has passed.
For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.
We can plug this values in our equations, to obtain:

note that the acceleration point downwards, hence the minus sign.
Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

and obtain:



Plugin this time on our first equation we find:

