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:
![V(t) = 38 \frac{m}{s} - g * t](https://tex.z-dn.net/?f=V%28t%29%20%3D%2038%20%5Cfrac%7Bm%7D%7Bs%7D%20-%20g%20%2A%20t)
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:
![0 m = 38 \frac{m}{s} - g * t](https://tex.z-dn.net/?f=%200%20m%20%3D%2038%20%5Cfrac%7Bm%7D%7Bs%7D%20-%20g%20%2A%20t)
and obtain:
![t = 38 \frac{m}{s} / g](https://tex.z-dn.net/?f=%20t%20%3D%2038%20%5Cfrac%7Bm%7D%7Bs%7D%20%2F%20g%20)
![t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}](https://tex.z-dn.net/?f=%20t%20%3D%2038%20%5Cfrac%7Bm%7D%7Bs%7D%20%2F%209.8%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%20)
![t = 3.9 s](https://tex.z-dn.net/?f=%20t%20%3D%203.9%20s%20)
Plugin this time on our first equation we find:
![y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2](https://tex.z-dn.net/?f=y%20%3D%2038%20%5Cfrac%7Bm%7D%7Bs%7D%20%2A%203.9%20s%20-%20%5Cfrac%7B1%7D%7B2%7D%209.8%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%20%283.9%20s%29%5E2)
![y=73.67 m](https://tex.z-dn.net/?f=y%3D73.67%20m)