C. Observations and measurements
B) Applied and gravitational forces
Answer: Part(a)=0.041 secs, Part(b)=0.041 secs
Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction
now we know that
a=-9.81 m/s^2 ( negative because it is pulling the player downwards)
we also know that
s=76 cm= 0.76 m ( maximum s)
using kinetic equation

where v is final velocity which is zero at max height and u is it initial
hence


now we can find time in the 15 cm ascent


using quadratic formula

t=0.0409 sec
the answer for the part b will be the same
To find the answer for the part b we can find the velocity at 15 cm height similarly using

where s=0.76-0.15
as the player has traveled the above distance to reach 15cm to the bottom


when the player reaches the bottom it has the same velocity with which it started which is 3.861
hence the time required to reach the bottom 15cm is

t=0.0409
The change in velocity from 30 m/s north to 40 m/s south is a change of 70 m/s south
Answer:
<em>a. The rock takes 2.02 seconds to hit the ground</em>
<em>b. The rock lands at 20,2 m from the base of the cliff</em>
Explanation:
Horizontal motion occurs when an object is thrown horizontally with an initial speed v from a height h above the ground. When it happens, the object moves through a curved path determined by gravity until it hits the ground.
The time taken by the object to hit the ground is calculated by:

The range is defined as the maximum horizontal distance traveled by the object and it can be calculated as follows:

The man is standing on the edge of the h=20 m cliff and throws a rock with a horizontal speed of v=10 m/s.
a,
The time taken by the rock to reach the ground is:


t = 2.02 s
The rock takes 2.02 seconds to hit the ground
b.
The range is calculated now:

d = 20.2 m
The rock lands at 20,2 m from the base of the cliff