Answer: The vertical displacement will be 127.5 meters
Step by step:
Since the question asks for the vertical displacement only, it is sufficient to limit our calculation to the vertical components of the problem.
We choose to measure the displacement on a vertical axis d with an initial coordinate 0 coinciding with the launch of the ball. There is an initial horizontal velocity v_0 of 50 m/s and the gravitational deceleration g acting in the direction opposite to the initial velocity. The formula for displacement of a body subject to an deceleration with an initial velocity (at time t=5 seconds) is:
[tex]d(t) = -\frac{1}{2}gt^2+v_0t+d_0=-\frac{1}{2}9.8 \frac{m}{s^2}t^2+50\frac{m}{s}\cdot t + 0m\\d(5) = -\frac{1}{2}9.8 \frac{m}{s^2}5^2 s^2+50\frac{m}{s}\cdot 5s=127.5m[/tex]
After 5 seconds of the launch, the ball will be at the point 127.5 meters from the origin on the vertical axis (vertical displacement)
