A ball is thrown up from a 10 -m-high cliff with an initial speed of 2.00 m/s. What is the speed of the ball just before it hits the ground? (The ground is defined as the bottom of the cliff.) Neglect air resistance
vf^2=vi^2 + 2*g*10
solve for vfinal.
V^2 = Vo^2 + 2g*h = 0.
4 + (-19.6)h = 0,
h = 0.20 m. above cliff.
ho + h = 10 + 0.2 = 10.2 m. above gnd.
V^2 = Vo^2 + 2g*(ho+h) = 0 + 19.6*10.2 = 200,
V = 14.14 m/s.
To find the speed of the ball just before it hits the ground, we need to use the laws of motion and the equations of kinematics.
First, we need to determine the initial velocity of the ball when it is thrown upwards. The problem states that the ball is thrown up with an initial speed of 2.00 m/s. Since the ball is thrown upwards, the initial velocity will be positive.
We also know that the height of the cliff is 10 meters, which is the maximum height the ball reaches before it starts falling back down.
Now, we can use the equation of motion to find the time it takes for the ball to reach the maximum height. The equation we can use is:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time taken. Since the ball is being thrown upwards against the force of gravity, the acceleration is -9.8 m/s^2 (taking downward as negative). The final velocity at the maximum height is 0 m/s, so we can rearrange the equation as:
0 = 2.00 m/s - 9.8 m/s^2 * t1
Solving for t1, we can find the time it takes for the ball to reach the maximum height.
Now, let's use another equation of motion to find the time it takes for the ball to fall from the maximum height to the ground. The equation we can use is:
h = vit + (1/2)at^2
where h is the height, vi is the initial velocity, a is the acceleration, and t is the time taken.
Since the height is 10 meters and the ball is falling downward with an acceleration of -9.8 m/s^2, we have:
10 m = 0 m/s * t2 + (1/2)(-9.8 m/s^2)(t2)^2
Now we can solve for t2, the time taken for the ball to fall from the maximum height to the ground.
Once we have both t1 and t2, we can find the total time taken by adding them up:
t_total = t1 + t2
Finally, to find the speed of the ball just before it hits the ground, we can use the equation of motion:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time taken. In this case, the initial velocity (vi) is 0 m/s (since the ball starts falling from rest), the acceleration (a) is -9.8 m/s^2, and the time taken (t) is t_total.
Substituting these values into the equation, we can solve for vf, which will be the speed of the ball just before it hits the ground.