Question

Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.95 m/s . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 39.6 m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 7.50 s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?

Answer 1

Answer:

Explanation:

Given

Speed of Henrietta is $v=2.95\ m/s$

Height of tower $h=39.6\ m$

Bruce throws the bangle after 7.5 s

During 7.5 s Henrietta travels

$x=2.95\times 7.5=22.125\ m$

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

$R=u\times t$

$R=x+2.95\times t-----1$

bangle will also cover a vertical distance of 39.6 m

so using equation of motion

$h=u_yt+\frac{1}{2}gt^2$

here initial vertical velocity is zero

$39.6=0+\frac{1}{2}\cdot 9.81\cdot t^2$

$t=\sqrt{8.0816}$

$t=2.84\ s$

Substitute the value of t in  

$u\times 2.84=22.125+2.95\times 2.84$

$u=10.743\ m/s$