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Thu 22 Sep, 2005 11:18 am
A ball has an initial velocity of 1.30 m/s along the +y-axis and, starting at t, receives an acceleration of 2.10 m/s^2 in the +x-direction.
a) What is the position of the ball 2.50 s after t?
b) What is the velocity of the ball at that time?
Treat the two forces as independent, that is if the axis perpendicular. The y factor is acceleration is independent of the x axis acceleration Vy is a constant 1.3 m/s.
The x axis is different. It x axis acceleration is handled a separate issue. If the initial x-axis velocity is 0 the x-axis velocity after 2.5 seconds of x-axis acceleration of 2.1 m/s^2 is (2.5)*(2.1) m/s or 5.25 m/s.
The total resultant velocity is the combination of these perpindiculear velocities.
@ 2.5 seconds
Vy=1.3 m/s
Vx=5.25 m/s
Total velocity
V=√(Vy^2+Vx^2)=√(1.3^2+5.25^2 )= =√(1.69+27.5625)= =√(29.2525)=5.408 m/s
The position is given by Sy=Vy*t=1.3m/s*2.5s=3.25 m in the y directions. The x-direction is given by Sx=1/2at^2=1/2*2.1*(2.5)^2=6.5625 m in the x-direction
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I'm so glad your intellect was able to help me out. You explained everything wonderfully! Thank you!
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