Question

a man named bungkas climbed an oak tree in 1970 and built himself a nest there. In 1994 he was still up there, and he has not left the tree for 24 years. suppos bungkas asks a villager for a newspaper, which is thrown to him straight up with an initial speed of 12 m/s. when bungkas catches the newspaper from his nest, the newspapers velocity is 3 m/s directed upward. from this information, find the height at which the nest was built. assumed the newspaper is thrown from a height 1.5 m above the ground.

Answer 1

To find the height of the nest where Bungkas is situated, we used a kinematic equation. By substituting the given values for initial and final velocity, and acceleration due to gravity, we calculated the height above the throwing point and added the height of throw from the ground, resulting in a total height of approximately 8.39 meters.

Step-by-step explanation:

The question presented involves a scenario where a newspaper is thrown straight up to Bungkas in a tree with an initial velocity of 12 m/s and is caught with a velocity of 3 m/s still directed upwards. Using physics formulas for kinematic equations, we can calculate the height at which the nest is built above the point of the throw.

To calculate the height, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v is the final velocity (3 m/s),
u is the initial velocity (12 m/s),
a is the acceleration (due to gravity, which is -9.8 m/s^2), and
s is the displacement (the height of the nest above the point of throw).

Plugging the values into the equation, we get:
(3 m/s)^2 = (12 m/s)^2 + 2(-9.8 m/s^2)s
9 = 144 - 19.6s
19.6s = 144 - 9
s = 135 / 19.6
s ≈ 6.89 m

This is the height of the nest above the point from where the newspaper is thrown. Since the newspaper is thrown from a height of 1.5 m above the ground, we add this to the displacement calculated:

Total height = 6.89 m + 1.5 m = 8.39 m (approximately)

Therefore, the nest was built approximately 8.39 meters above the ground.