Answer:
The initial velocity of the softball is 14.711 meters per second.
Explanation:
This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.
From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:
(Eq. 1)
Where:
- Initial height of the softball, measured in meters.
- Final height of the softball, measured in meters.
- Initial velocity of the softball, measured in meters per second.
- Time, measured in seconds.
- Gravitational acceleration, measured in meters per square second.
If we know that 
, 
and 
, the initial velocity of the softball is:
The initial velocity of the softball is 14.711 meters per second.
Answer:
V is greater
Explanation:
because v intial at that time V final is the that speed which it is going at that time
It is the branch of science, in which we study different phenomena of atmosphere including climate and weather.
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = 
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
Answer:
Over such small distances, digital data may be transmitted as direct, two-level electrical signals over simple copper conductors. This results from the electrical distortion of signals traveling through long conductors, and from noise added to the signal as it propagates through a transmission medium.
