The answer i think is (B)
Answer:
a) the distance the ball moves up the ramp is approximately 9.35 meters
b) the tme required for the ball to return to the girls' hands is 6.34 seconds
Explanation:
The movement of the ball can be describe by a rectilinear movement with constant (negative) acceleration of 0.19 g = 1.862 m/s^2
Therefore the kinematic equations for the case the girl launches the ball at 5.9 m/s, we can write:
We use the second equation above to find the time it takes for the velocity to reduce to zero (and start the movement back down the ramp), and then use the time found to calculate the distance;
We use this time to find the distance the ball moves up the ramp:
so the distance the ball moves up the ramp is approximately 9.35 meters
The ball will return to the hands of the girl in the double of the time it took to reach the max distance, that is 2 times 3.17 sec = 6.34 sec
Answer:
density I believe, air gets compressed when In pressure, so the air would have more density
Answer:
It would take the object 5.4 s to reach the ground.
Explanation:
Hi there!
The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:
h = h0 + v0 · t + 1/2 · g · t²
Where:
h = height of the object at time t.
h0 = initial height.
v0 = initial velocity.
g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).
t = time
Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:
h = h0 + 1/2 · g · t²
We have to find the time at which h = 0:
0 = 470 ft - 1/2 · 32.2 ft/s² · t²
Solving for t:
-470 ft = -16.1 ft/s² · t²
-470 ft / -16.1 ft/s² = t²
t = 5.4 s
Answer:
B- Velocity
Explanation:
This means gravity makes the Moon accelerate all the time, even though its speed remains constant.