An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?
To determine the initial (launch) speed of the object, we can use the laws of projectile motion.
The maximum height of an object thrown vertically can be calculated using the equation:
h_max = (v_initial)^2 / (2 * g)
where:
h_max is the maximum height
v_initial is the initial velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Given that the object reaches one fourth of its maximum height above its launch point, we can express this as:
h = h_max / 4
Substituting the values into the equation:
h_max / 4 = (v_initial)^2 / (2 * g)
Now, let's solve for the initial velocity (v_initial):
Multiply both sides of the equation by (2 * g):
h_max / 4 * (2 * g) = (v_initial)^2
Simplify:
h_max / 2 * g = (v_initial)^2
Now, take the square root of both sides of the equation:
sqrt(h_max / 2 * g) = v_initial
Since we have the value of g (9.8 m/s^2) and the object reaches one-fourth of its maximum height, we can substitute these values into the equation:
sqrt(h_max/2 * 9.8) = v_initial
Finally, we can calculate the initial velocity (v_initial) by evaluating the right side of the equation using a calculator.
To determine the initial (launch) speed of the object, we can use the concept of conservation of mechanical energy.
Let's break down the problem step by step:
Step 1: Identify the given information:
- The upward velocity of the object (v) when it reaches one fourth (1/4) of its maximum height above its launch point: v = 18 m/s
- The distance at which the velocity is given (one fourth of the maximum height)
Step 2: Understand the concept of conservation of mechanical energy:
Conservation of mechanical energy states that the sum of kinetic energy (KE) and potential energy (PE) remains constant throughout the motion of the object in the absence of external forces such as air resistance.
Step 3: Utilize the concept of conservation of mechanical energy:
When the object is at one fourth of its maximum height, the kinetic energy is equal to the potential energy. Mathematically, this can be expressed as:
(1/2)mv^2 = mgh/4
Where:
m is the mass of the object (which is not given in this problem but cancels out)
v is the upward velocity of the object
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the maximum height reached by the object
We can simplify the equation by canceling out the mass (m) on both sides:
(1/2)v^2 = gh/4
Step 4: Solve for the initial (launch) speed:
To find the initial (launch) speed, we need to determine the maximum height (h) reached.
Since the object is thrown vertically upwards, it will eventually reach a maximum height where its upward velocity becomes zero before it starts descending.
At the maximum height, the object's velocity is zero, so we have:
v = 0
Using the equation:
v^2 = u^2 + 2gh
where u is the initial (launch) speed,
0 = u^2 + 2gh
From this equation, we can find the value of gh:
gh = -u^2/2
Now, let's substitute this value into the previous equation derived from the conservation of mechanical energy:
(1/2)v^2 = gh/4
(1/2)(18^2) = (-u^2/2)/4
9^2 = -u^2/8
81 = -u^2/8
-u^2 = 8 * 81
u^2 = -8 * 81
To continue, we need to take the square root of both sides:
u = sqrt(-8 * 81)
However, we cannot take the square root of a negative number in this context. It means our assumption was incorrect. Therefore, there is no real solution for the initial (launch) speed of the object given the information provided.
16.5306
v = Vo - 9.8 t
height versus time is a parabola
h = Vo t - 4.9 t^2
Vertex is at the top where v = 0 call that height capital H
0 = Vo - 9.8 ttop
ttop = Vo/9.8
H = Vo ttop - 4.9 ttop^2
H = Vo^2/9.8 - .5 Vo^2/9.8
H = .5 (Vo^2/9.8)
H/4 = .125 (Vo^2/9.8)
at some time t it is at H/4 and v =18
v = 18 = Vo - 9.8 t
H/4 = Vo t - 4.9 t^2 = .125 Vo^2/9.8
Vo = 18 + 9.8 t
that gives two equations in the two unknowns, t and Vo
solve for Vo