Answer:
25.3 m/s
Explanation:
First, we need to calculate the time that it takes to reach the ground, so we will use the following equation
[tex]y=v_{iy}t+\frac{1}{2}gt^2[/tex]
Where Viy is the initial vertical velocity, so Viy = 0 m/s, y is the height of 22m and g is the gravity, so g = 10 m/s². Replacing the values and solving for t, we get
[tex]\begin{gathered} 22=0t+\frac{1}{2}(10)t^2 \\ 22=5t^2 \\ \frac{22}{5}=t^2 \\ \\ 4.4=t^2 \\ \sqrt{4.4}=t \\ 2.1\text{ s = t} \end{gathered}[/tex]
Then, with the time t = 2.12 s and the horizontal distance R = 53 m, we can calculate the initial horizontal velocity as
[tex]\begin{gathered} v=\frac{distance}{time} \\ \\ v=\frac{53\text{ m}}{2.1\text{ s}} \\ \\ v=25.26\text{ m/s} \end{gathered}[/tex]
Therefore, the answer is 25.3 m/s
