A juggler throws a ball from height of 0.950 m with a vertical velocity of +5.50 m/s and misses it on the way down. What is its velocity when it hits the ground?
Vf = Vo + gt = 0,
5.5 + (-9.8)t = 0,
5.5 - 9.8t = 0,
9.8t = 5.5,
t = 5.5 / 9.8 = 0.56 s = time to reach max. height.
d = Vo - 4.9t^2,
d = 5.5 - 4.9(0.56)^2,
d = 5.5 - 1.54 = 3.96 m = max. height reached.
d(tot.) = 3.96 + 0.95 = 4.91 m above
gnd.
Vf^2 = Vo^2 + 2gd,
Vf^2 = o + 2 * 9.8 * 4.91,
Vf^2 = 96.3,
Vf = 9.81 m. = Velocity when it hits the gnd.
can you please help me with this identical problem?
A juggler throws a ball from height of 0.950 m with a vertical velocity of +4.25 m/s and misses it on the way down. What is its velocity when it hits the ground?
To find the velocity of the ball when it hits the ground, we can use the principle of conservation of energy.
When the ball is thrown upwards, it gains potential energy due to its height. As it comes down, it loses potential energy and gains an equal amount of kinetic energy. At the same time, the ball's initial upward velocity will gradually decrease due to the force of gravity pulling it downwards.
To solve this problem, we first need to calculate the initial potential energy of the ball. We can use the formula:
Potential Energy (PE) = mass (m) * gravity (g) * height (h)
In this case, the initial height is 0.950 m, and the mass of the ball is not given. However, mass does not affect the calculation of potential energy since it cancels out when we calculate the velocity. So we can ignore it for now.
Next, we need to calculate the final velocity when the ball hits the ground. This can be done using the equation for the conservation of energy:
Potential Energy (PE) = Kinetic Energy (KE)
The potential energy at the top, when the ball is thrown, is:
PE(top) = m * g * h
The kinetic energy at the bottom, when the ball hits the ground, is:
KE(bottom) = (1/2) * m * v^2
We know that the initial vertical velocity is +5.50 m/s. In this case, we only need the magnitude of the velocity, so we'll use the positive value.
Since the ball is thrown from a height of 0.950 m, it will go up and then come back down to the ground. At the top of its trajectory, the ball's velocity will be 0 m/s. Therefore, the final kinetic energy (KE) will be 0.
Setting the potential energy at the top equal to the kinetic energy at the bottom, we get:
m * g * h = (1/2) * m * v^2
We can cancel out the mass (m) on both sides of the equation, giving us:
g * h = (1/2) * v^2
To solve for the velocity (v), we can rearrange the equation:
v^2 = 2 * g * h
Finally, we take the square root of both sides to find the magnitude of the velocity when the ball hits the ground:
v = sqrt(2 * g * h)
Using the values for the acceleration due to gravity (g = 9.8 m/s^2) and the height of 0.950 m, we can now calculate the velocity:
v = sqrt(2 * 9.8 * 0.950) = 4.19 m/s
Therefore, the velocity of the ball when it hits the ground is approximately 4.19 m/s.