Answer:
Explanation:
Given
Speed of Henrietta is [tex]v=2.95\ m/s[/tex]
Height of tower [tex]h=39.6\ m[/tex]
Bruce throws the bangle after 7.5 s
During 7.5 s Henrietta travels
[tex]x=2.95\times 7.5=22.125\ m[/tex]
Suppose bangle hit the ground after t sec so bangle will has to cover a distance of x and distance traveled by Henrietta during this time t
Range of bangle when thrown with speed u
[tex]R=u\times t[/tex]
[tex]R=x+2.95\times t-----1[/tex]
bangle will also cover a vertical distance of 39.6 m
so using equation of motion
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
here initial vertical velocity is zero
[tex]39.6=0+\frac{1}{2}\cdot 9.81\cdot t^2[/tex]
[tex]t=\sqrt{8.0816}[/tex]
[tex]t=2.84\ s[/tex]
Substitute the value of t in
[tex]u\times 2.84=22.125+2.95\times 2.84[/tex]
[tex]u=10.743\ m/s[/tex]