Question

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

Answer 1

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 $\sqrt{(2d)/a}$. Solving the equation where d=8 and a=9.8 makes the time $\sqrt{(2*8)/9.8}$=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.

Answer 2

Answer:

The answer is 12.5 m/s

v=the square root of 2 x(gh)

or v = the square root of 2 x (9.8 x 8)

Explanation: