A monkey in a perch 20 m high in a tree drops a coconut above the head of a zoo keeper as he runs with a speed 1.5 m/s beneath the tree actually intending to hit the toes of the zoo keeper, how early in seconds should the coconut be dropped by the monkey
time to fall= sqrt (2h/g)
distance to run=1.5*time
1.5 m
To determine how early the coconut should be dropped, we need to consider the time it takes for the coconut to fall from the perch to the height of the zookeeper's toes.
First, let's calculate the time it takes for the coconut to fall from the perch to the ground. We can use the following formula:
h = (1/2) * g * t^2
where:
h is the height (20 m),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
and t is the time.
Plugging in the values, we get:
20 = (1/2) * 9.8 * t^2
Simplifying the equation:
40 = 9.8 * t^2
Dividing both sides by 9.8:
t^2 = 40 / 9.8
Taking the square root of both sides:
t ≈ √(40 / 9.8) ≈ √4.08 ≈ 2.02 seconds
So, it takes approximately 2.02 seconds for the coconut to fall from the perch to the ground.
Now, we need to determine how early the coconut should be dropped to hit the toes of the zookeeper. We want the coconut to hit the zookeeper at the same time they reach the height of their toes, which is 20 m below the perch.
Since the zookeeper is running beneath the tree with a speed of 1.5 m/s, we can calculate the distance they cover in 2.02 seconds:
distance = speed * time = 1.5 * 2.02 ≈ 3.03 m
Therefore, the coconut should be dropped 3.03 meters above the zookeeper's toes to hit them at the right time.
To calculate how early in seconds the coconut should be dropped, we can use the following equation:
time = distance / speed = 3.03 / 1.5 ≈ 2.02 seconds
So, the coconut should be dropped 2.02 seconds early to hit the zookeeper's toes.