Two trees have perfectly straight trunks and are both growing perpendicular to the flat horizontal ground beneath them. The sides of the trunks that face each other are separated by 1 m. A frisky squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that is 0.9 m above the ground on the other tree. Then, he jumps back to the first tree, landing on it at a spot that is 1.8 m above the ground. Finally, he leaps to the other tree, now landing at a spot that is 2 m above the ground. What is the magnitude of the squirrel's displacement?
To find the magnitude of the squirrel's displacement, we need to calculate the distance between the starting point and the ending point of the squirrel's jumps.
Let's break down the problem into steps:
1. The squirrel leaps from the foot of one tree to a spot that is 0.9 m above the ground on the other tree. In this step, the squirrel moves upward by 0.9 m, but there is no horizontal displacement. So, we can ignore this step when calculating the final displacement.
2. The squirrel jumps back to the first tree and lands at a spot that is 1.8 m above the ground. In this step, the squirrel moves upward by an additional 0.9 m (1.8 m - 0.9 m), and there is still no horizontal displacement.
3. Finally, the squirrel leaps to the other tree and lands at a spot that is 2 m above the ground. In this step, the squirrel moves upward by an additional 0.2 m (2 m - 1.8 m), and again, there is no horizontal displacement.
To calculate the total displacement, we need to add up the vertical displacements:
0.9 m + 0.9 m + 0.2 m = 2 m
The magnitude of the squirrel's displacement is 2 meters.
h^2=1.3^2+2.5^2=1.69+6.25=7.94
h=sqrt7.94
=2.82m