A golf ball is released from rest from the top of a very tall building. Choose a coordinate system whose origin is at the starting point of the ball, and whose y axis points vertically upward.
The acceleration of gravity is 9.8 m/s^2.
a.Neglecting air resistance, calculate the velocity of the ball after 2.04 s.
Answer in units of m/s.
b.What is the position of the ball after 2.04 s? (Neglect air resistance.)
Answer in units of m.
Use Henry's Equation, however make your answer negative since gravity is -9.8m/s^2
a. Well, it seems like the golf ball is in a bit of a hurry! To calculate its velocity after 2.04 seconds, we can use the equation for motion under constant acceleration. The initial velocity is zero since the ball is released from rest, and the acceleration is -9.8 m/s^2 (negative because it's going downward due to gravity). So, plugging these numbers into the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is time, we get:
v = 0 + (-9.8) * 2.04
After some quick calculations, the velocity of the ball after 2.04 seconds comes out to be approximately -20 m/s. But wait, don't worry, the negative sign just means the velocity is directed downward, which is only natural for a falling object like our golf ball.
b. Alright, let's see where our speedy little golf ball ends up after 2.04 seconds. Again, we can utilize the equation for motion under constant acceleration, but this time we'll use the equation s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is time.
Since the initial velocity is zero and the acceleration remains the same at -9.8 m/s^2, let's plug in the values:
s = 0 * 2.04 + (1/2) * (-9.8) * (2.04)^2
Let's do some more number-crunching and we find that the position of the ball after 2.04 seconds is approximately -20.08 meters. Now, that's what I call a downward trajectory! Just remember to look on the bright side - at least the ball won't have to worry about hitting any birds on its way down!
a. To find the velocity of the ball after 2.04 s, we can use the following equation:
v = u + a*t
where:
v is the final velocity
u is the initial velocity (which is 0 m/s since the ball is released from rest)
a is the acceleration due to gravity (9.8 m/s^2)
t is the time (2.04 s)
Plugging in the values, we get:
v = 0 + 9.8 * 2.04
v ≈ 19.992 m/s
Therefore, the velocity of the ball after 2.04 s is approximately 19.992 m/s.
b. To find the position of the ball after 2.04 s, we can use the following equation:
s = u*t + (1/2)*a*t^2
where:
s is the displacement or position
u is the initial velocity (0 m/s)
t is the time (2.04 s)
a is the acceleration due to gravity (9.8 m/s^2)
Plugging in the values, we get:
s = 0 * 2.04 + (1/2) * 9.8 * (2.04)^2
s ≈ 20.004 m
Therefore, the position of the ball after 2.04 s is approximately 20.004 m.
To answer these questions, we can use the equations of motion for an object in free fall. The key equation to use is:
y = y0 + v0*t + (1/2)*a*t^2
Where:
- y is the final position of the object
- y0 is the initial position of the object
- v0 is the initial velocity of the object
- a is the acceleration of the object (in this case, the acceleration due to gravity)
- t is the time elapsed
Let's calculate the velocity of the ball after 2.04 seconds first.
a. To find the velocity, we need to know the initial velocity of the ball. Since the golf ball is released from rest, the initial velocity is 0 m/s.
Plugging these values into the equation, we get:
y = y0 + v0*t + (1/2)*a*t^2
y = 0 + 0*2.04 + (1/2) * 9.8 * (2.04)^2
Simplifying, we have:
y = 0 + 0 + (1/2)*9.8*(4.1616)
y = 0 + 0 + 20.40528
y ≈ 20.41 m
So, after 2.04 seconds, the position of the ball is approximately 20.41 meters.
b. To find the position of the ball after 2.04 seconds, we can again use the same equation as above.
y = y0 + v0*t + (1/2)*a*t^2
y = 0 + 0*2.04 + (1/2) * 9.8 * (2.04)^2
Plugging in the values, we have:
y = 0 + 0 + (1/2)*9.8*(4.1616)
y = 0 + 0 + 20.40528
y ≈ 20.41 m
So, after 2.04 seconds, the position of the ball is approximately 20.41 meters above the starting point.
a. V = Vo + g*t = 0 + 9.8*2.04 = 20 m/s
b. d = 0.5g*t^2 = 4.9*2.04^2 = 20.39 m.
Downward.