derive the formula
v0=sqrt((deltaX)squared)*g)/2*deltaY
To derive the formula, let's start with the equation of motion for an object that is dropped vertically:
deltaY = (1/2) * g * t^2 -- Equation 1
Here,
- deltaY represents the change in height (or displacement) of the object.
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
- t is the time it takes for the object to fall.
We can rearrange Equation 1 to solve for t:
t^2 = (2 * deltaY) / g -- Equation 2
Next, let's consider the horizontal motion of the object. We assume that there is no horizontal acceleration, so the velocity of the object in the horizontal direction will be constant. The horizontal distance covered by the object can be calculated using the formula:
deltaX = v0 * t -- Equation 3
Here,
- deltaX represents the horizontal displacement.
- v0 is the initial velocity of the object in the horizontal direction.
- t is the time.
Now we substitute the value of t from Equation 2 into Equation 3:
deltaX = v0 * sqrt((2 * deltaY) / g)
To solve for v0, we can rearrange the equation:
v0 = deltaX / sqrt((2 * deltaY) / g)
Finally, to simplify the equation further, we can multiply both the numerator and denominator by sqrt(g):
v0 = (deltaX * sqrt(g)) / sqrt(2 * deltaY)
And this is the derived formula for v0 in terms of deltaX and deltaY:
v0 = (deltaX * sqrt(g)) / sqrt(2 * deltaY)