Answer:
-5.29 m/s
Explanation:
Given:
y₀ = 1.43 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: v
v² = v₀² + 2a (y − y₀)
v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.43 m)
v = -5.29 m/s
Answer
given,
range of the projectile = 4.3 m
time of flight = T = 0.829 s
v = 5.19 m/s
vertical component of velocity of projectile
v_y = gt'
a) Launch angle
θ = 38°
b) initial speed of projectile
v = 6.59 m/s
c) maximum height reached by the projectile
Answer:
As they vibrate, they pass the energy of the disturbance to the particles next to them, which pass the energy to the particles next to them, and so on. Particles of the medium don't actually travel along with the wave. Only the energy of the wave travels through the medium
<span>Remember that impulse = change in
momentum
this means we compute the momentum of the ball just before impression and just
after; we know the mass, so we find the speeds
the ball falls for 1.5m and will achieve a speed given by energy
conservation:
1/2 mv^2 = mgh => v=sqrt[2gh]=5.42m/s
since it rises only to 0.85 m, we compute the initial speed after power from
the same equation and get
v(after)=sqrt[2*9.81m/s/s*0.85m] = 4.0837 m...
now, recall that momentum is a vector, so that the momentum down has one sign and
the momentum up has a positive sign, so we have
impulse = delta (mv) = m delta v = 0.014 kx (4.08m/s - (-5.42m/s) = 0.133 kgm/s </span>
Calculated Speed for the entire trip can be called as "Average Speed".
