Answer:
= -32.53 m / s
this velocity is directed downwards
Explanation:
This is a free fall exercise, let's use the expression
= v_{oy}^{2} + 2 g (y -yo)
where we are assuming that there is friction with the air, as the body falls its initial velocity is zero
v_{oy} = √ 2g (y - y₀)
let's calculate
v_{y} = √ (2 9.8 (0-54.0))
= -32.53 m / s
this velocity is directed downwards
Answer:
(a): The car's relative position to the base of the cliff is x= 32.52m.
(b): The lenght of the car in the ir is tfall= 1.78 sec.
Explanation:
Vo= 0
V= ?
d= 50m
h= 30m
a= 4 m/s²
t= √(2*d/a)
t= 5 sec
V= a*t
V= 20 m/s
Vx= V * cos(24º)
Vx= 18.27 m/s
Vy= V* sin(24º)
Vy= 8.13 m/s
h= Vy*t + g*t²/2
clearing t:
tfall= 1.78 sec (b)
x= Vx * tfall
x= 32.52 m (a)
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When something is hit harder just like when sound is turned up the waves become higher and more frequent like a zig zag more so then wavy.