Stephanie serves a volleyball from a height of
0.78 m and gives it an initial velocity of +7.7
m/s straight up.
How high will the volleyball go? The acceleration of gravity is 9.81 m/s^2.
How long will it take the ball to reach its
maximum height?
Well, Stephanie really gave that volleyball a lift, didn't she? Let's find out how high it will go and how long it will take.
To calculate the maximum height, we can use the formula:
h = (v^2 - u^2) / (2g),
where h is the height, v is the final velocity (which would be 0 m/s at the highest point), u is the initial velocity (7.7 m/s), and g is the acceleration due to gravity (9.81 m/s^2).
Plugging in the values, we get:
h = (0^2 - 7.7^2) / (2 * 9.81)
or h = -59.29 / 19.62
which gives us h = -3.02 m.
Oh no, it seems we've hit negative heights! But don't worry, it just means we've chosen the upwards direction as positive, so the height would be positive. Therefore, the volleyball will go to approximately 3.02 meters high.
Now, let's move on to the time it takes to reach the maximum height.
We can use the formula:
t = (v - u) / g,
where t is the time, v is the final velocity at maximum height (0 m/s), u is the initial velocity (7.7 m/s), and g is the acceleration due to gravity (9.81 m/s^2).
Plugging in the values, we get:
t = (0 - 7.7) / 9.81,
or t = -7.7 / 9.81,
which gives us t = -0.785 seconds.
Wait a minute, negative time? That's not possible! Well, it's just a sign convention issue. We assumed upwards as positive, so the time turned out to be negative. In reality, it would take approximately 0.785 seconds for the volleyball to reach its maximum height.
So there you have it. The volleyball will go up to around 3.02 meters high, and it will take about 0.785 seconds to reach that height. Keep serving, Stephanie!
To find the maximum height reached by the volleyball, we can use the kinematic equation:
vf^2 = vi^2 + 2ad
Where:
- vf is the final velocity (which will be 0 at the maximum height)
- vi is the initial velocity (+7.7 m/s)
- a is the acceleration (-9.81 m/s^2)
- d is the displacement or the maximum height we want to find
Rearranging the equation to solve for d:
d = (vf^2 - vi^2) / (2a)
Substituting the values:
d = (0 - (7.7 m/s)^2) / (2 * -9.81 m/s^2)
Simplifying:
d = (-59.29 m^2/s^2) / (-19.62 m/s^2)
d = 3.02 m
Therefore, the volleyball will reach a height of 3.02 meters.
To find the time it takes for the ball to reach its maximum height, we can use the equation:
vf = vi + at
Where:
- vf is the final velocity (which will be 0 at the maximum height)
- vi is the initial velocity (+7.7 m/s)
- a is the acceleration (-9.81 m/s^2)
- t is the time we want to find
Substituting the values:
0 = 7.7 m/s + (-9.81 m/s^2) * t
Solving for t:
-7.7 m/s = -9.81 m/s^2 * t
t = (-7.7 m/s) / (-9.81 m/s^2)
t = 0.785 seconds
Therefore, it will take the ball approximately 0.785 seconds to reach its maximum height.
To determine how high the volleyball will go, we can use the equation for the vertical motion of an object:
v_f^2 = v_i^2 + 2aΔy
Here,
v_f = final velocity (which will be 0 at the maximum height),
v_i = initial velocity,
a = acceleration due to gravity (-9.81 m/s^2 since the ball is moving upward),
Δy = change in vertical position (which is the height of the ball).
Solving for Δy, we have:
Δy = (v_f^2 - v_i^2) / (2a)
Substituting the given values:
v_i = +7.7 m/s
v_f = 0 m/s
a = -9.81 m/s^2
Δy = (0^2 - 7.7^2) / (2 * -9.81)
= (-59.29) / (-19.62)
= 3.02 meters
Therefore, the volleyball will reach a height of 3.02 meters.
Now, let's find the time it takes for the ball to reach its maximum height. We can use the equation for the vertical motion of an object:
v_f = v_i + at
Here,
v_f = final velocity (which will be 0 at the maximum height),
v_i = initial velocity,
a = acceleration due to gravity (-9.81 m/s^2),
t = time taken.
Solving for t, we have:
t = (v_f - v_i) / a
Substituting the given values:
v_i = +7.7 m/s
v_f = 0 m/s
a = -9.81 m/s^2
t = (0 - 7.7) / (-9.81)
= 0.785 seconds
Therefore, it will take the ball approximately 0.785 seconds to reach its maximum height.
It travels up untils its vertical velocity is zero. That happens after t = 7.7/g = 0.785 seconds. (That answers your second question). During that time, it moves upward
t*Vaverage =
(1/2)*(7.7)*(0.785) = 3.02 m
Add that to the height from which it was released, and you get 3.80 meters