Answer:
[tex]u=14.48m/s[/tex]
Explanation:
From the question we are told that:
Height of window [tex]h=2m[/tex]
Height of window off the ground [tex]h_g=7.5m[/tex]
Time to fall and drop [tex]t=1.3s[/tex]
Generally the Newton's equation motion is mathematically given by
[tex]s=ut+\frac{1}{2}at^2[/tex]
Where
[tex]h=ut+\frac{1}{2}at^2[/tex]
[tex]2=u1.3-\frac{1}{2}*9.8*1.3^2[/tex]
[tex]2=u1.3-8.281[/tex]
[tex]u=7.91m/s^2[/tex]
Generally the Newton's equation motion is mathematically given by
[tex]2as=v^2-u^2[/tex]
Where
[tex]-2gh_g=v^2-u^2[/tex]
[tex]-2*9.8*7.5=(7.91)^2-u^2[/tex]
[tex]-147=62.5681-u^2[/tex]
[tex]u=\sqrt{209.5681}[/tex]
[tex]u=14.48m/s[/tex]
Therefore the ball’s initial speed
[tex]u=14.48m/s[/tex]
