A particle is dropped from the height h and falls freely for a time,t.With the aid of a sketch, explain how h varies with. a. t b. t^2
the height y(t) = h - 1/2 gt^2
Your sketch should show a parabola with vertex at (0,h) and x-intercept at (ā(2h/g),0)
the 2nd sketch is just a straight line through (0,h) and (2h/g,0)
I need explanation for question above
To understand how the height, "h," varies with time, "t," when a particle is dropped and falls freely, we can refer to the laws of motion and use a simple kinematic equation.
a. Relationship between h and t:
When we drop a particle and it falls freely, it experiences acceleration due to gravity. This acceleration is constant and acts downward. As a result, the particle's velocity increases over time, causing it to cover more distance (height) during each subsequent time interval.
To represent the relationship between h and t, we can use a sketch of a height-time graph. The graph will have time, "t," on the x-axis and height, "h," on the y-axis.
At the initial time, t=0, the particle is released from height "h" and starts falling. As time passes, the height "h" decreases. The relationship between "h" and "t" is linear, following a downward-sloping line.
At any given time, "t," the height "h" can be calculated using the equation for free fall motion:
h = h0 - (1/2)gt^2
Where:
- h0: Initial height (height from which the particle was dropped)
- g: Acceleration due to gravity (approximately 9.8 m/sĀ²)
- t: Time elapsed
b. Relationship between h and t^2:
If we want to study the relationship between h and t^2, we can rewrite the equation mentioned above by isolating t:
t = ā(2(h0 - h) / g)
Now, let's substitute this value of t into the equation and observe how h varies with t^2:
h = h0 - (1/2)g((ā(2(h0 - h) / g))^2)
Simplifying further:
h = h0 - (1/2)g(2(h0 - h) / g)
h = h0 - (h0 - h)
h = h - h0
Therefore, we can see that "h" is equal to "-h0" or simply a constant value. Hence, the relationship between "h" and "t^2" is that "h" remains constant (assuming h0 is constant).
Note: The value of "h" would only change with respect to "t" and not "t^2" when considering free fall motion.