Question

Henrietta is jogging on the side-walk at 3.05 m/s on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is 38.0 m above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally 9.00 s after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?

Answer 1

Answer:

12.9121148614 m/s

35.9393048982 m

Explanation:

t = Time taken

u = Initial velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$s=ut+\frac{1}{2}at^2\\\Rightarrow 38=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{38\times 2}{9.81}}\\\Rightarrow t=2.78337865516\ s$

Time taken for the bag to fall is 2.78337865516 seconds

Time she has been jogging for

9+2.78337865516 = 11.78337865516 seconds

Total distance traveled by her

$s=vt\\\Rightarrow s=3.05\times 11.78337865516=35.9393048982\ m$

Henrietta is 35.9393048982 m away

Velocity of throwing

$\dfrac{35.9393048982}{2.78337865516}=12.9121148614\ m/s$

The velocity of throwing is 12.9121148614 m/s