Boy throws a penny into a wishing well at a velocity of 2 m/s and at an angle of 45 degrees below horizontal. If the penny takes 1.3 seconds to hit the water, how deep is the well?
To determine how deep the well is, we can use the kinematic equations of motion. Firstly, let's break down the given information:
- Initial velocity of the penny, vo = 2 m/s
- Launch angle below horizontal, θ = 45 degrees
- Time taken for the penny to hit the water, t = 1.3 seconds
Now, we need to split the initial velocity into horizontal and vertical components. The horizontal component, vx, remains constant throughout the motion, while the vertical component, vy, changes due to the acceleration due to gravity.
To find the horizontal component of velocity, we can use the formula:
vx = vo * cos(θ)
Substituting the values, we have:
vx = 2 * cos(45°) = 2 * √2 / 2 = √2 m/s
The vertical component of velocity can be determined using:
vy = vo * sin(θ)
Substituting the values, we have:
vy = 2 * sin(45°) = 2 * √2 / 2 = √2 m/s
Since the penny is thrown downwards, the initial vertical velocity, vy, should be negative. So, vy = -√2 m/s.
Next, let's determine the time of flight, which is the total time the penny is in the air. We know that the time of flight is twice the time taken to reach the maximum height:
t_flight = 2 * t = 2 * 1.3 seconds = 2.6 seconds
Now, using the time of flight and the vertical component of velocity, we can find the maximum height reached by the penny. The formula is:
y_max = (vy^2) / (2 * g)
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values, we have:
y_max = (-√2^2) / (2 * 9.8) = -2 / 19.6 = -0.102 m
The negative sign indicates that the maximum height is below the initial position.
Finally, we can calculate the depth of the well. The depth of the well is the sum of the maximum height and the initial height:
Depth = y_max + initial height
Since we assume the initial height is zero (the penny is launched from ground level), the depth of the well is simply the magnitude of the maximum height:
Depth = |y_max| = |-0.102 m| = 0.102 m
Therefore, the well has a depth of approximately 0.102 meters.
Vo = 2m/s[-45o]
Xo=2*cos(-45)=1.414 m/s=Hor. component.
Yo=2*sin(-45)=-1.414 m/s=Ver. component.
Depth = Yo*t + 0.5g*t^2 =
-1.414*1.3 + 4.9*1.3^2 = 6.44 m.