A squirrel sees a BIG acorn being thrown from a tree by a chipmunk and attempts to snatch it. The acorn is thrown from a 10.0 m high branch and with a horizontal velocity of 1.1 m/s. The squirrel on the ground is 5.0 m from the base of the tree. What is the of the squirrel’s average velocity if it grabs the acorn in its mouth just as it touches the ground?
Can someone explain this?
So he throws it horizontal.
time in air: 10=1/2 a t^2
t=sqrt(20/9.8) seconds
how far will the acorn travel: 1.1*sqrt(20/9.8)
how far does the squirrel run?
1.1*sqrt(20/9.8) -5
average velocity squirrel ?
distance above/timeto fall
Travel = 1.571429
Distance the squirrel ran = −3.428571
Is this right?
I would take numbers to more than two sig digits. Now if the squirrel is on the other side of the tree, distance will be 5+1.6 m. The problem is not clear where the squirrel is.
To solve this problem, we need to determine the time it takes for the acorn to reach the ground and then calculate the squirrel's average velocity.
First, let's find the time it takes for the acorn to fall to the ground. We can use the basic kinematic equation for vertical motion:
h = (1/2) * g * t^2
Where:
- h is the height of the tree branch (10.0 m in this case),
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time it takes for the acorn to fall.
Rearranging the equation to solve for t, we have:
t^2 = (2 * h) / g
t^2 = (2 * 10.0) / 9.8
t^2 ≈ 2.04
Taking the square root of both sides, we find that t ≈ 1.43 seconds.
Now we can find the horizontal distance the acorn will travel during this time. We're given that the horizontal velocity of the acorn is 1.1 m/s, and the time is 1.43 seconds, so:
distance = velocity * time
distance ≈ 1.1 * 1.43
distance ≈ 1.573 m
The squirrel is 5.0 m from the base of the tree, and the distance the acorn traveled horizontally is 1.573 m. This means the squirrel will have to move an additional distance to catch the acorn. Therefore, the squirrel's average velocity can be calculated as:
average velocity = total distance / total time
The total distance is 5.0 m + 1.573 m = 6.573 m. And the total time is the same as the time it took for the acorn to fall, which is 1.43 seconds.
Therefore, the average velocity of the squirrel as it catches the acorn is:
average velocity = 6.573 m / 1.43 s
average velocity ≈ 4.59 m/s
So, the squirrel's average velocity, when it grabs the acorn in its mouth just as it touches the ground, is approximately 4.59 m/s.