A water balloon is thrown straight down at 10.0 m/s from a second floor window, 5.00 m above ground level.
1) How fast is the balloon moving when it hits the ground? (Express your answer to three significant figures.)
v^2 = vo^2 - 2g(y-yo)
v^2 = 10^2 - 2(9.8)(0-5)
v = √198
v = 14.1 m/s
Well, isn't this a "splash-tastic" question! Let's dive right in.
To solve this, we can use the kinematic equation that relates the final velocity (vf), initial velocity (vi), acceleration (a), and displacement (d):
vf^2 = vi^2 + 2ad
Since the water balloon is thrown straight down, the initial velocity (vi) is 10.0 m/s and acceleration (a) is -9.8 m/s^2 (negative because it's going downwards). The displacement (d) is 5.00 m.
Now, let's plug in the values and calculate:
vf^2 = (10.0 m/s)^2 + 2(-9.8 m/s^2)(5.00 m)
vf^2 = 100 m^2/s^2 - 98 m^2/s^2
vf^2 = 2 m^2/s^2
Taking the square root of both sides, we find:
vf = sqrt(2 m^2/s^2) ≈ 1.41 m/s
So, when the water balloon hits the ground, "water" you know, it's moving at approximately 1.41 m/s. Keep dry and stay punny!
To find the speed of the water balloon when it hits the ground, we can use the equation for motion under constant acceleration:
v^2 = u^2 + 2as
Where:
v = final velocity (unknown)
u = initial velocity = 10.0 m/s
a = acceleration = 9.81 m/s^2 (due to gravity)
s = displacement = -5.00 m (since the balloon is moving downward)
Substituting the known values into the equation:
v^2 = (10.0 m/s)^2 + 2 * 9.81 m/s^2 * (-5.00 m)
v^2 = 100 m^2/s^2 + (-98.1 m^2/s^2) * (-5.00 m)
v^2 = 100 m^2/s^2 - 490.5 m^2/s^2
v^2 = -390.5 m^2/s^2
Taking the square root of both sides:
v = √(-390.5 m^2/s^2)
Since velocity cannot be negative in this context (magnitude represents speed), we discard the negative sign:
v ≈ 19.8 m/s
Therefore, the water balloon is moving at approximately 19.8 m/s when it hits the ground.
To solve this problem, we can use the equation of motion that relates the initial velocity, final velocity, acceleration, and displacement:
v^2 = u^2 + 2as
Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
s is the displacement.
Let's assign values to the variables:
u = 10.0 m/s (initial velocity)
a = 9.8 m/s^2 (acceleration due to gravity, assuming downward direction)
s = 5.00 m (displacement)
Now, let's plug in the values into the equation:
v^2 = (10.0 m/s)^2 + 2 * 9.8 m/s^2 * 5.00 m
v^2 = 100.0 m^2/s^2 + 98.0 m^2/s^2 * 5.00 m
v^2 = 100.0 m^2/s^2 + 490.0 m^2/s^2
v^2 = 590.0 m^2/s^2
To find v, we take the square root of both sides:
v = √(590.0 m^2/s^2)
Now, we can calculate the value of v using a calculator or approximating as follows:
v ≈ √(600 m^2/s^2) ≈ 24.5 m/s
So, the balloon is moving at approximately 24.5 m/s when it hits the ground.
h = -1/2 g t^2 - 10.0 t + 5.00
find the time (T) when h equals zero (impact)
v = 10.0 + g T