Answer:
The maximum height reached by the body is 313.6 m
The time to return to its point of projection is 8 s.
Explanation:
Given;
initial velocity of the body, u = 78.4 m/s
at maximum height (h) the final velocity of the body (v) = 0
The following equation is applied to determine the maximum height reached by the body;
v² = u² - 2gh
0 = u² - 2gh
2gh = u²
h = u²/2g
h = (78.4²) / (2 x 9.8)
h = 313.6 m
The time to return to its point of projection is calculated as follows;
at maximum height, the final velocity becomes the initial velocity = 0
h = v + ¹/₂gt²
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ = 
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt = ![\frac{1}{g} \ [(v_{o} - v) - ( - v_{o} - v) ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bg%7D%20%5C%20%5B%28v_%7Bo%7D%20-%20v%29%20%20-%20%28%20-%20v_%7Bo%7D%20%20-%20v%29%20%5D)
Δt = 2 v₀ / g
Known variables
d=4.6m
initial velocity=0m/s
downward acceleration=-9.8m/s2
d=1/2gt2
4.6=1/2 -9.8 t2
t=0.93s
C is the right image for that biological process.
One charge is enough in order to have an electric field.
In fact, every charge generates an electric field: for example, the electric field generated by a single point positive charge is radial, as shown in the attached figure. More complicate electric field configurations can be obtained adding charges or using more complicate charge distributions, but one charge is enough to have an electric field.
