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?
First, we need to find the time it takes for the bagels to hit the ground after Bruce throws them.
To do this, we can use the equation for vertical motion:
Δy = v₀t + (1/2)gt²
where:
Δy = -43.9 m (negative because it is in the opposite direction of the positive direction)
v₀ = 0 (initial vertical velocity, bagels are thrown horizontally)
t = time (unknown)
g = 9.8 m/s² (acceleration due to gravity)
Plugging in the values, we get:
-43.9 = 0 * t + (1/2) * 9.8 * t²
Simplifying the equation, we have:
-43.9 = (4.9)t²
Next, we solve for t:
t² = -43.9 / 4.9
t² ≈ -8.98
Since time cannot be negative, we discard the negative solution.
Therefore, t ≈ √(8.98)
t ≈ 2.997 s (approximately)
So, it takes approximately 2.997 seconds for the bagels to hit the ground after Bruce throws them.
Now, to find the initial speed at which Bruce must throw the bagels, we can use the horizontal distance covered by Henrietta during this time.
The horizontal distance Henrietta covers while the bagels are in the air is:
Distance = (Henrietta's speed) * (Time taken by the bagels to hit the ground)
Distance = 2.05 m/s * 2.997 s
Distance ≈ 6.1417 m (approximately)
Bruce needs to throw the bagels with an initial speed that allows them to travel this distance before hitting the ground.
Since there is no acceleration in the horizontal direction, the equation we use is:
Distance = (Initial horizontal velocity) * (Time taken by the bagels to hit the ground)
6.1417 m = (Initial horizontal velocity) * 2.997 s
Solving for the Initial horizontal velocity, we get:
Initial horizontal velocity ≈ 6.1417 m / 2.997 s
Initial horizontal velocity ≈ 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 principles of projectile motion. The bagels will have an initial horizontal velocity and a vertical velocity due to gravity.
First, let's analyze the vertical motion of the bagels. The window is 43.9 m above the ground, and it takes Henrietta 9.00 s to reach the same vertical position. We can use the equation of motion for vertical motion:
y = vt - (1/2)gt^2
where:
y is the vertical displacement (43.9 m)
v is the initial vertical velocity (unknown)
t is the time (9.00 s)
g is the acceleration due to gravity (9.8 m/s^2)
Plugging in the values, we can solve for the initial vertical velocity:
43.9 = v(9.00) - (1/2)(9.8)(9.00)^2
Simplifying the equation, we find:
43.9 = 9v - 441
Rearranging the terms:
9v = 43.9 +441
9v = 484.9
v = 484.9 / 9
v ≈ 53.87 m/s
So, the initial vertical velocity of the bagels must be approximately 53.87 m/s.
Now let's consider the horizontal motion. Since there is no horizontal force acting on the bagels, their horizontal velocity will remain constant throughout the motion. Henrietta is moving with a speed of 2.05 m/s, so Bruce must throw the bagels with a horizontal velocity equal to Henrietta's speed.
Therefore, the initial horizontal velocity of the bagels thrown by Bruce must be 2.05 m/s.
In conclusion, to throw the bagels at the right angle for Henrietta to catch them, Bruce needs to throw them with an initial velocity of approximately 2.05 m/s horizontally and 53.87 m/s vertically.