- #1

R.Dahl

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*Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.5 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 56.0 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 cath them just before they hit the ground?

Where is Henrietta when she catches the bagels?

With what initial speed must Bruce throw the bagels so Henrietta can cath them just before they hit the ground?

Where is Henrietta when she catches the bagels?

First I determined how far down the sidewalk Henrieta would be by d=vt (d=2.5*7.5=18.75).

Then I used the equation [y=y_0 + v_0y - 1/2*g*t^2], substituting into get [0=56 - 4.91*t^2], I found time t to be 3.38s,

and finally using the equation [x=x_0 + v_0x*t] I substituted [18.75 = v_0x*3.38] and solved for v_0x as 5.55 m/s. I am at a loss as to what I did wrong.

Secondly for the second part of the problem which asks how far from the building the woman is when she catches the bagel would that answer be found by d=v*t, based on my understanding of the problem we are basing the speed of the bagel on how far it needs to be thrown before striking the ground.

I've reworked it several times but it keeps telling me the answer is wrong.