Asked by arlene
A ball is thrown upward. Its initial vertical component of velosity is 30m/s and its initial horizontal component of velosity is 20m/s. What is the ball's speed 2 s later? Give: Viy = 30m/s Vix = 20m/s T = 2 sec
a) V = gt =10m/sec2 x 2 sec= 20 m/sec
b) c2 = (20m/s)2 + (20m/s)2 = 400m2/s2 +400m2/2 = 800m2/s2 c = 28.28 =Vf V =gt = 10m/s2(2 sec) = 5m/s
a) V = gt =10m/sec2 x 2 sec= 20 m/sec
b) c2 = (20m/s)2 + (20m/s)2 = 400m2/s2 +400m2/2 = 800m2/s2 c = 28.28 =Vf V =gt = 10m/s2(2 sec) = 5m/s
Answers
Answered by
Henry
Xo = 20 m/s
Yo = 30 m/s
Vo = sqrt(20^2+30^2) = 36.1 m/s
Y = Yo + g*t = 30 + (-9.8)*2 = 10.4 m/s.
= Ver. component of velocity 2 s later.
X = Xo = 20 m/s and remains constant.
V^2 = X^2 + Y^2 = 20^2 + 10.4^2 = 508.16
V = 22.5 m/s. = Total velocity.
Yo = 30 m/s
Vo = sqrt(20^2+30^2) = 36.1 m/s
Y = Yo + g*t = 30 + (-9.8)*2 = 10.4 m/s.
= Ver. component of velocity 2 s later.
X = Xo = 20 m/s and remains constant.
V^2 = X^2 + Y^2 = 20^2 + 10.4^2 = 508.16
V = 22.5 m/s. = Total velocity.
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