Explanation:
The maximum height of an object, given the initial launch angle and initial velocity is found with:h=v2isin2θi2g h = v i 2 sin 2 θ i 2 g .
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Answer: 244.05 J
Explanation:
To find speed at 30 m above the ground use equation:
V²=Vo²-2Gs
V0=31.4m/s
s=30m
G=9.81m/s²
-----------------------
V²=31.4²-2*9.81*30
V²=985.96+588.6
V²=1574.56
V=39.68m/s ---speed of arrow on 30 m obove the ground
Use equation for kinetic enrgy:
Ke=mV²/2
m=0.155kg
V=39.68m/s
-------------------------
Ke=0.155kg*(39.68m/s)²/2
Ke=0.155*1574.5/2
Ke=244.05J
Explanation:
PRIMERO HACES EL RECUENTO DEL TIEMPO Y LO CONVIERTES EN
SEGUNDOS Y ENTONCES
<em>t</em> = 227 s 
= 227 S - 38 s = 189 s
= 38 s
LUEGO USANDO LA ECUACIÓN DE GALILEO GALILEI SSUPONIENDO
QUE EL MOVIL VIAJA A VELOCIDAD CONSTANTE
<em>v</em> = 3.50 m/189 s = 0.0185 m/s
PARA LA DISTANCIA NTRE B Y C
= 0.0185 m/S( 38 s) = 0.703 m
LA HORA EN QUE EL MOVIL PASA POR A ES
11:43:15 - 38 s - 189 s = 11:39:29
As the greater force of tension (by 81N) is exerted by the team on the right the rope will move to the right.
