Aball player catches a ball 3.13 s after throwing it vertically upward. a)with what speed did he throw it?
b)what height did it reach?
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Ответ:
a) v₀ = 15.3 m/s
b) h = 11.9 m
Explanation:
Ball Kinematics
We apply the free fall formulas:
vf= v₀+at Formula (1)
vf²=v₀²+2*a*y Formula (2)
y= v₀t+ (1/2)*a*t² Formula (3)
y:displacement in meters (m)
v₀: initial speed in m/s
f: final speed in m/s
a: acceleration in m/s²
Data
t= 3.13 s
a=g= -9.8 m/s² = acceleration due to gravity
Problem development
a)With what speed did he throw it
t= 3.13 s : total ball time going up and downThe time the ball rises to its maximum height is half the total time At maximum height vf = 0We apply the formula (1) for calculate initial speed (v₀)
vf= v₀+at
0= v₀+(-9.8)*(3.13/2)
v₀= 15.3 m/s
b)What height did it reach?
We apply formula (2) or formula (3)
vf²=v₀²+2*a*h formula (2)
0 = v₀² +2*(-9.8)*h
19.6*h = v₀²
h=v₀²/19.6
h= (15.3)²/(19.6 )
h= 11.9m
y= v₀t+ (1/2)*a*t² Formula (3)
h= (15.3)(3.13/2)+(1/2)(-9.8)(3.13/2)²= 11.9m
Ответ: