A small steel ball bearing with a mass of 21.0 g is on a shortcompressed spring.

Question

A small steel ball bearing with a mass of 21.0 g is on a shortcompressed spring. When aimed vertically and suddenly released, thespring sends the bearing to a height of 1.21 m. Calculate thehorizontal distance the ball would travel if the same spring wereaimed 29.0 deg from the horizontal.
Consider the horizontal and vertical components of velocityseperately. Can someone give me the numerical answer? Consider the horizontal and vertical components of velocityseperately. Can someone give me the numerical answer? Consider the horizontal and vertical components of velocityseperately.

Explanation / Answer

m=21.0g=0.0210kg let init speed of the bearing when is vertical is v, wehave: (1/2)mv2 = mgh =>v =2gh=(2*9.8m/s2*1.21m)=4.87m/s when it release at angle 29.0 degree, since the releasepotential energy is the same so it result the same release speed v=4.87m/s but not vertical. vx = 4.87 m/s *cos(29.0) = 4.26 m/s vy = 4.87m/s *sin(29.0) = 2.36 m/s the travel time is t = 2vy / g = 0.48s the horizontal distance is: D= vx*t = 4.26m/s *0.48s = 2.05 m .

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