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Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.05 m/sm/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 40.8 mm above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 7.00 ss after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.

Required:
a. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
b. Where is Henrietta when she catches the bagels?

Respuesta :

Answer:

v = 10.46 m/s

x =  30.134 m from house

Explanation:

given data

speed = 3.05 m/s

time = 7 s

height = 40.8 mm

solution

we get here first time required to fall that is t

t = [tex]\sqrt{\frac{2\times 40.8}{9.8}}[/tex]       ..................1

t = 2.88 s

now we take here initial speed that is v

so v × t = 3.05 × ( t+ 7)

v = [tex]\frac{3.05\times (2.88 + 7)}{2.88}[/tex]

v = 10.46 m/s

and

when she catch the bagel henrietta was at

x = 3.05 × ( 2.88 + 7)

x =  30.134 m from house

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