Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 2.05 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 43.9 m above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 9.00 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?
h = Vo*t + 0.5g*T^2 = 43.9 m,
0 + 4.9t^2 = 43.9,
t^2 = 8.96,
Tf = 3.0 s. = Time to fall to gnd.
d = r*t = 2.05m/s * 9s = 18.45 m. = Dist. from window to jogger.
Vo*Tf = 18.45 m.
Vo*3 = 18.45,
Vo = 18.45 / 3 = 6.15 m/s.
To find the initial speed at which Bruce must throw the bagels, we can use the kinematic equations of motion. First, let's break down the problem and identify the known variables:
Henrietta's speed (v1) = 2.05 m/s
Time elapsed before Bruce throws the bagels (t) = 9.00 s
Height of the window above the sidewalk (h) = 43.9 m
We can consider the horizontal and vertical motions of the bagels separately.
Vertical Motion:
The bagels will fall from a height of 43.9 m to the ground, so we'll use the equation:
h = (1/2) * g * t^2
where:
h is the height of the window above the sidewalk
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time it takes for the bagels to fall (same as the time Bruce threw them after Henrietta has passed below the window)
Rearranging the equation to solve for t:
t = sqrt(2h / g)
Substituting the values, we get:
t = sqrt(2 * 43.9 / 9.8)
t ≈ 3.02 s
Horizontal Motion:
The distance Henrietta runs (d) during this time can be calculated using the equation:
d = v1 * t
Substituting the values, we get:
d = 2.05 * 3.02
d ≈ 6.18 m
Now, we can find the initial horizontal velocity (v2) with which Bruce must throw the bagels so that they will land at the same position as Henrietta:
d = v2 * t
Rearranging the equation to solve for v2:
v2 = d / t
Substituting the values, we get:
v2 = 6.18 / 3.02
v2 ≈ 2.05 m/s
Therefore, Bruce must throw the bagels with an initial horizontal velocity of approximately 2.05 m/s for Henrietta to catch them just before they hit the ground.
To solve this problem, we can use the equations of motion in the horizontal and vertical directions separately.
First, let's consider the horizontal motion. The horizontal velocity of the bagels remains constant throughout their flight since there is no horizontal acceleration. We are given that Henrietta runs at a speed of 2.05 m/s, so the horizontal velocity of the bagels is also 2.05 m/s.
Next, let's analyze the vertical motion. We can use the equation for vertical displacement:
Δy = Viy * t + (1/2) * g * t^2,
where Δy represents the vertical displacement, Viy represents the initial vertical velocity, t represents the time, and g represents the acceleration due to gravity (approximately 9.8 m/s^2).
In this case, the vertical displacement, Δy, is equal to the height of the window, which is 43.9 m. Since the bagels are thrown horizontally, the initial vertical velocity, Viy, is 0.
So the equation becomes:
43.9 = 0 * t + (1/2) * 9.8 * t^2.
Simplifying this equation gives:
4.9 * t^2 = 43.9.
Dividing both sides by 4.9 gives:
t^2 = 8.98.
Taking the square root of both sides gives:
t ≈ 2.99 s.
Now we can use the horizontal velocity and the time to find the horizontal distance traveled by the bagels:
Δx = Vx * t,
where Δx represents the horizontal distance and Vx represents the horizontal velocity.
Substituting the values, we have:
Δx = 2.05 m/s * 2.99 s ≈ 6.12 m.
Therefore, Bruce must throw the bagels with an initial horizontal speed of approximately 6.12 m/s for Henrietta to catch them just before they hit the ground.