You throw a ball downward from a window at a speed of 2.0m/s. Theball accelerate

Question

You throw a ball downward from a window at a speed of 2.0m/s. Theball accelerates at 9.8 m/s2. How fast is it moving when it hitsthe sidewalk 2.5 m below. I believe the answer that I want willcome about by using the acceleration formula, however, after that,I'm not sure which is what, and how to use the accelerationformula.

Explanation / Answer

Hi. In fact, there are 2 "acceleration formulas": x = xo + vot +(1/2)at2 v = vo + at Here xo and vo are the initial velocities. Inproblems, they are often set to zero. (But not here...) To solve the problem, you'll have to use both. Look at the firstequation I wrote. Compare it to the problem statement. What ismissing? We have a distance, an initial velocity (speed), anacceleration. What is missing is the time. Once you have the timeelapsed, you will be able to solve the 2nd equation I gaveyou. This equation will tell you how fast it was moving whenit hit the sidewalk (i.e., at the end of the elapsed time youfound). So, now that we know what to do, all we need to do is a little bitof algebra. First, let xo = 0. I'll rewrite the equation in a form you will recognise: (a /2) t2 + vo t - x = 0 All I did is "switch x to the other side" and change the order ofthe terms. Why did I do that? So that you could find the similaritywith the equation:     a x2 + b x  + c = 0      -------->    x = [ -b +(b2-4ac) ] / 2a   and    [-b- (b2-4ac) ] / 2a You know how to solve this equation for x, do you? Now youhave to solve the preceding equation, but for t. Only one ofthe two solutions should make sense. Pick this one. Once you have the value for t, plug it in the second equationabove: v = vo + at = (2.0 m/s) + (9.8m/s2) t And now you know how fast the ball was moving when it hit thesidewalk.

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