Bert is standing on a ladder picking apples in his father's orchard. As he pulls each apple off the tree, he tosses it into a basket that sits on the ground 2.2 m below at a horizontal distance of 1.3 m from Bert. How fast must Bert throw the apples (horizontally) in order for them to land in the basket? Enter m/s as unit and use g = 10. m/s2.
To find the required horizontal velocity at which Bert must throw the apples, we can use the equations of motion.
First, let's find the time it takes for the apples to reach the ground. We can use the vertical motion equation:
y = ut + (1/2)gt^2
Where:
y = vertical displacement (2.2 m, negative since it's downwards)
u = initial vertical velocity (0 m/s, since the apple starts at rest)
g = acceleration due to gravity (-10 m/s^2)
t = time taken
Plugging in the values, we get:
-2.2 = 0t + (1/2)(-10)(t^2)
-2.2 = -5t^2
t^2 = 0.44
t = sqrt(0.44)
t ≈ 0.663 s
Since the horizontal distance is given as 1.3 m, and the time taken to reach the ground is 0.663 s, we can now calculate the horizontal velocity using the equation:
v = d/t
Where:
v = horizontal velocity
d = horizontal distance (1.3 m)
t = time taken (0.663 s)
Plugging in the values, we get:
v = 1.3 / 0.663
v ≈ 1.963 m/s
Therefore, Bert must throw the apples horizontally with a velocity of approximately 1.963 m/s for them to land in the basket.
To determine how fast Bert must throw the apples horizontally, we need to consider the horizontal motion and vertical motion of the apple.
Let's start by analyzing the vertical motion of the apple. The apple falls vertically under the influence of gravity. We can use the equations of motion to find the time the apple takes to fall from the tree to the basket.
The distance the apple falls vertically is 2.2 m, and the acceleration due to gravity is 10 m/s^2. We can use the equation:
h = (1/2)gt^2
Where:
h is the vertical distance (2.2 m),
g is the acceleration due to gravity (10 m/s^2), and
t is the time taken.
Rearranging the equation to solve for t, we get:
t^2 = (2h) / g
t^2 = (2 x 2.2) / 10
t^2 = 0.44 / 10
t^2 = 0.044
t ≈ 0.21 s
Now that we know the time it takes for the apple to fall vertically to the ground, we can determine the horizontal distance the apple travels during this time. We know that the horizontal distance is 1.3 m.
Since the horizontal motion is uniform (no net force acts in the horizontal direction), the horizontal velocity remains constant throughout the motion. We can use the equation:
d = v*t
Where:
d is the horizontal distance (1.3 m),
v is the horizontal velocity, and
t is the time taken (0.21 s).
Rearranging the equation to solve for v, we get:
v = d / t
v = 1.3 / 0.21
v ≈ 6.19 m/s
Therefore, Bert must throw the apples horizontally with a speed of approximately 6.19 m/s for them to land in the basket.
while falling, the apple obeys
2.2 = 10t^2
so it hits the ground after 0.66 sec.
To travel horizontally 1.3m in 0.66 sec, it must travel
1.3m/.66s = 1.97m/s
Naturally, there is some wiggle room, depending on the size of the basket.