Speed of the arrow just before it will hit the ground can be calculated by energy conservation
by solving above equation
now this will lodge itself 15 cm into the ground
so after lodging itself to 15 cm the speed of arrow becomes zero
now we will have
Now by Newton's II law net force is given as
Data given:
V=45m/s
t=11s
Δx=?
Formula needed:
V=Δx/t
Solution:
Δx=v×t
Δx=45m/s×11s
Δx=495m
According to my solution B) is the most accurate
Answer:
Plato, Aristotle developed it further and used for 1400 years till Copernicus.
Explanation:
Answer:
A) 19.994 m B) 1.750 seconds
Explanation:
We are asked about time and height at which both balls pass at the same height. So both, <em>TIME </em>and <em>HEIGHT</em> of both vertical trajectories are the same.
To find the values we use the kinematics expression for vertical motion with constant acceleration, using as the gravity acceleration -9.8 m/s2. Motion downwards is negative and upwards is positive. The reference point is the bottom of the building. The equation is as follow:
For each ball this equation is:
Since the data of the problem is using SI units, then our answers will be expressed in SI units as well. Now, we first compute part B by equaling both equations (same height) and solving for time:
With this time we can find the Height by substituting it in any equation of the balls. In this case, we use the expression of ball B