Playing jackstone as a kid, you discovered that there is a certain time before the rubber ball falls and you took advantage of this when picking up the stars on the floor. If you threw the ball 2.72 m/s up into the air, how fast will it be moving at a height of 48.80 cm? How much time has elapsed upon reaching this height?
How long does it take to reach 48.8cm?
2.27t - 4.9t^2 = 0.488
Now you have the time, and
v = 2.72 - 9.81t
To answer this question, we need to consider the motion of the ball thrown up into the air. We can use the principles of projectile motion to find the answers.
The velocity of the ball at any height can be determined using the equation:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
In this case, we want to find the final velocity (v) of the ball at a height of 48.80 cm (or 0.488 m), knowing the initial velocity (u) is 2.72 m/s and the acceleration (a) is -9.8 m/s^2 (assuming downward as negative).
Let's plug the values into the equation:
v^2 = u^2 + 2as
Substituting the values:
v^2 = (2.72 m/s)^2 + 2(-9.8 m/s^2)(0.488 m)
Solving the equation, we get:
v^2 = 7.3984 m^2/s^2 + (-9.6056 m^2/s^2)
v^2 = -2.2072 m^2/s^2
Since we are only interested in the magnitude of the velocity, we can take the square root of both sides of the equation:
v = √(-2.2072 m^2/s^2)
However, remember that the ball is moving upward, so the resulting velocity will be positive. Let's take the positive square root:
v = √(2.2072 m^2/s^2)
Calculating the square root, we find the velocity (v) to be approximately 1.486 m/s at a height of 48.80 cm.
Next, we can calculate the time it takes for the ball to reach this height.
The equation for displacement in vertical motion is:
s = ut + (1/2)at^2
where:
s = displacement (0.488 m)
u = initial velocity (2.72 m/s)
a = acceleration (-9.8 m/s^2)
t = time
Plugging in the values, we have:
0.488 m = (2.72 m/s)t + (1/2)(-9.8 m/s^2)t^2
Now, we can rearrange the equation to solve for time (t). This results in a quadratic equation, which we can solve using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
In our case, the quadratic equation is:
(-4.9 m/s^2)t^2 + (2.72 m/s)t - 0.488 m = 0
Comparing this equation to the quadratic formula (ax^2 + bx + c = 0), we have:
a = -4.9 m/s^2
b = 2.72 m/s
c = -0.488 m
Now, we can substitute these values into the quadratic formula:
t = [-(2.72 m/s) ± √((2.72 m/s)^2 - 4(-4.9 m/s^2)(-0.488 m))] / 2(-4.9 m/s^2)
Simplifying further, we have:
t = [-(2.72 m/s) ± √(7.3984 m^2/s^2 + 4.7952 m^2/s^2)] / (-9.8 m/s^2)
t = [-(2.72 m/s) ± √(12.1936 m^2/s^2)] / (-9.8 m/s^2)
Taking the positive time value, we have:
t ≈ [-(2.72 m/s) + √(12.1936 m^2/s^2)] / (-9.8 m/s^2)
Calculating the square root and evaluating the expression, we find the time (t) to be approximately 0.376 seconds.
Therefore, the ball will be moving at a speed of approximately 1.486 m/s and it takes approximately 0.376 seconds to reach a height of 48.80 cm.