A monkey throws a banana horizontally from the top of a tree. The banana hits the ground 3.03 seconds later and lands d = 30.9 meters from the base of the tree.
1. What was the resultant velocity of the banana as it hit the ground?
magnitude = ? m/sec.
2. What is the angle ?° from the horizontal?
horizontal speed is constant: 30.9/3.03 m/s
The vertical speed is v = at = -4.9t
The angle x is such that tan(x) = Vy/Vx
Oops -- v = -9.8t
To solve the problem, we can use the kinematic equations of motion. We'll start by finding the initial velocity of the banana.
Given:
Time, t = 3.03 s
Horizontal displacement, d = 30.9 m
1. Finding the initial velocity:
Use the equation: d = vt
Rearranging the equation to solve for the initial velocity (v):
v = d/t
Substituting the given values into the equation:
v = 30.9 m / 3.03 s
v ≈ 10.199 m/s
So, the magnitude of the resultant velocity of the banana as it hits the ground is approximately 10.199 m/s.
2. Finding the angle from the horizontal:
Since the banana is thrown horizontally, the angle from the horizontal is 0 degrees.
Therefore, the angle is 0° from the horizontal.
To find the resultant velocity of the banana as it hits the ground, we can use the equation of motion for horizontal motion, since the banana is thrown horizontally.
1. The equation of motion for horizontal motion is: d = v*t, where d is the distance traveled horizontally, v is the horizontal velocity, and t is the time taken.
Given that the banana lands at d = 30.9 meters from the base of the tree and t = 3.03 seconds, we can rearrange the equation to solve for v:
v = d/t
v = 30.9 meters / 3.03 seconds
v ≈ 10.198 m/s
So, the magnitude of the resultant velocity of the banana as it hits the ground is approximately 10.198 m/s.
2. Since the banana is thrown horizontally, the angle from the horizontal is 0°.