Physics
RegineMiernik
24

A ball of a mass 0.3 kg is released from rest at a height of 8 m. How fast is it going when it hits the ground? Acceleration due to gravity is g=9.8 m/s^2

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(1) Answers
karinaazzahs

In order to solve this problem, there are two equations that you need to know to solve this problem and pretty much all of kinematics. The first is that d=0.5at^2 (d=vertical distance, a=acceleration due to gravity and t=time). The second is vf-vi=at (vf=final velocity, vi=initial velocity, a=acceleration due to gravity, t=time). So to find the time that the ball traveled, isolate the t-variable from the d=0.5at^2. Isolate the t and the equation now becomes [latex] \sqrt{(2d)/a} [/latex]. Solving the equation where d=8 and a=9.8 makes the time [latex] \sqrt{(2*8)/9.8} [/latex]=1.355 seconds. With the second equation, the vi=0 m/s, the vf is unknown, a=9.8 m/s^2 and t=1.355 sec. Substitute all these values into the equation vf-vi=at, this makes it vf-0=9.8(1.355). This means that the vf=13.28 m/s.

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