a boy on a small hill aims his water ballon slingshot horizontally straight horizontally at a second boy hanging from a branch of a tree at a distance of d away . at the instant ballon is released the second boy let go and fall off froom tree to aviod eing hit will his plan work? why or why not?
If the dropping boy is far enough away that the balloon cannot reach him, he won't be hit.
Otherwise, since he falls at the same rate as the balloon, he will be hit as he drops.
Well, isn't that a slippery situation! If the first boy is aiming his water balloon slingshot horizontally, and the second boy lets go of the branch and falls just as the balloon is released, the second boy might just avoid getting hit. Why? Because the horizontal motion of the water balloon will continue uninterrupted, while the second boy's vertical fall will be independent of the balloon's horizontal path. So, if the timing is just right, the second boy might just make a splash-free getaway! But remember, timing is everything, and it's a "fall or fly" situation. Good luck, gravity-defying second boy!
To determine whether the second boy's plan will work or not, we need to consider the motion of the water balloon and the falling boy.
When the boy on the hill releases the water balloon, it will travel horizontally due to the initial velocity given by the slingshot. This means that the horizontal distance traveled by the water balloon will not be affected by the effect of gravity.
Meanwhile, the second boy is falling freely due to the force of gravity. The time it takes for him to fall from the branch to the ground can be calculated using the equation for the time of free fall:
t = sqrt(2f/g)
Where:
- t represents the time of free fall
- f represents the height from which the boy fell
- g represents the acceleration due to gravity (approximately 9.8 m/s²)
In this case, the height from which the boy fell is not mentioned, but we can assume it is small compared to the horizontal distance d. Therefore, we can neglect the time taken by the boy to fall.
Since the horizontal distance traveled by the water balloon does not change due to gravity and the falling boy will not have enough time to reach the ground, the water balloon will hit the boy as intended. Thus, the second boy's plan will not work, and he will be hit by the water balloon.
To determine whether the second boy's plan will work, we need to consider the motion of the water balloon and the falling boy. Let's break down the problem:
1. Projectile Motion of the Water Balloon: When the boy releases the water balloon, it will follow a projectile motion. This means that the water balloon will follow a curved path due to the combination of its horizontal and vertical velocities. However, since the question states that the water balloon is aimed "horizontally straight," it means that the initial vertical velocity of the water balloon is zero. Consequently, the water balloon will only travel horizontally.
2. Falling Boy: The second boy falls from the branch at the exact moment the water balloon is released. Since the water balloon is only traveling horizontally, its path will not intersect the falling boy's vertical motion. This is because the water balloon continues moving at a constant horizontal velocity while the falling boy accelerates vertically due to gravity.
Therefore, the falling boy's plan will work. By falling straight down from the branch, he will avoid being hit by the horizontally moving water balloon.