Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.65m/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 54.8m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 4.50s after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.

Question

Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.65m/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 54.8m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 4.50s 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

d = r * t = 2.65m/s * 4.5s = 11.9 m =

Distance ran by Henrietta in 4.5s.

h = 0.5g*t^2 = 54.8 m.
4.9t^2 = 54.8
t^2 = 11.18
Tf = 3.34 s. = Fall time.

d = Vo*Tf = 11.9 m.
Vo*3.34 = 11.9
Vo = 3.56 m/s.

Answer 2

To find the initial speed at which Bruce must throw the bagels, we can use the kinematic equations of motion. Let's break down the problem step by step.

Step 1: Calculate the time it takes for the bagels to reach the ground after Bruce throws them.

To calculate the time it takes for the bagels to reach the ground, we can use the equation:

h = (1/2) * g * t^2

Where:
h = height of the window above the ground (54.8m)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time it takes for the bagels to reach the ground

Solving for t:

t^2 = (2h) / g
t^2 = (2 * 54.8) / 9.8
t^2 = 5.6
t = sqrt(5.6)
t ≈ 2.37 s

Step 2: Calculate the horizontal distance Henrietta covers in 4.50s.

Since Henrietta is jogging horizontally at a constant speed, the horizontal distance she covers in 4.50s is given by:

d = v * t

Where:
v = Henrietta's jogging speed (2.65 m/s)
t = time interval (4.50s)

d = 2.65 * 4.50
d ≈ 11.93m

Step 3: Calculate the horizontal distance the bagels need to travel in 2.37s.

Since the bagels need to reach Henrietta just before they hit the ground, the horizontal distance they need to travel in 2.37 seconds is the same as the distance she covered in 4.50 seconds.

Therefore, the horizontal distance the bagels need to travel is approximately 11.93m.

Step 4: Calculate the initial horizontal velocity of the bagels.

To find the initial horizontal velocity at which Bruce must throw the bagels, we can use the equation:

d = v0 * t

Where:
d = horizontal distance (11.93m)
v0 = initial horizontal velocity (unknown)
t = time (2.37s)

Solving for v0:

v0 = d / t
v0 = 11.93 / 2.37
v0 ≈ 5.03 m/s

So, Bruce must throw the bagels with an initial horizontal velocity of approximately 5.03 m/s for Henrietta to catch them just before they hit the ground.