We use a fundamental kinematic equation as follows:
V = Vo + g*t.
<span>Tr = (V-Vo)/g = (0-10)/-10 = 1 s. = </span><span>time to reach max. height </span>
<span>Tf = Tr = 1 s. = Fall time or time to fall back to edge of bldg. </span>
<span>3-Tr-Tf = 3-1-1 = 1 s. Below edge of bldg. </span>
<span>d = Vo*t + 0.5g*t^2. </span>
<span>d = 10*1 + 5*1^2 = 15 m. <---- OPTION C</span>
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 =
Δt = 2 v₀ / g
Answer:
They would attract one another
Explanation:
The interaction between two like-charged objects is repulsive. ... Positively charged objects and neutral objects attract each other negatively charged objects and neutral objects attract each other.
Answer:
just guys
Explanation:
and if not i need how old you are sorry just trying to be safe
momentum= mass × velocity
p= 50×18
momentum= 900 kg m/s