Answer:
73.67 m
Explanation:
If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:
[tex]y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2[/tex]
[tex]V(t) = V_0 + a * t[/tex],
Where [tex]y_0[/tex] its our initial height, [tex]V_0[/tex] 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:
[tex]y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2[/tex]
[tex]V(t) = 38 \frac{m}{s} - g * t[/tex]
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:
[tex] 0 m = 38 \frac{m}{s} - g * t[/tex]
and obtain:
[tex] t = 38 \frac{m}{s} / g [/tex]
[tex] t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2} [/tex]
[tex] t = 3.9 s [/tex]
Plugin this time on our first equation we find:
[tex]y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2[/tex]
[tex]y=73.67 m[/tex]
