A) Please explain the steps and why I use the equation that you will use to deri

Question

A) Please explain the steps and why I use the equation that you will use to derive the answer please!

I know the answer is Vf= .5 *g * t

I just need to know what equation is used to derive this, why it is used, and the steps to getting to this answer.

b)

2- If the time the ball remains in the air is t, calculate the maximum height hmax the ball attained while in the air.

I know the answer is #5... but how do I derive that?

A ball is thrown upward with an initial vertical velocity of v0 to a maximum height of h max and falls back toward Earth. On the way down, it is caught at the same height at which it was thrown upward. If the total time the ball remains in the air is t, find its speed when caught. The acceleration of gravity is g. Neglect air resistance.

Explanation / Answer

using v=u+a*t

for going up we have u=Vo, v=0 a=-g

so we get t1=Vo/g for going up

again we use v=u+a*t for going down... u=o, v is not known a = g and t is also not known ..

using v^2-u^2 = 2*a*s for the whole trip gives initial velocity of the trip equals final velocity..since displacement is zero

so for coming down we have u=0,v=Vo,a=g, t is not known

now usong v=u+a*t => t2 = Vo/g so total trip time is t=t1+t2 = 2*Vo/g => t=2*Vo/g

since velocity when the ball is caught is t we have Vo= 1/2*g*t =.5g*t

use v^2-u^2 = 2*a*s for way up///

v=0, u = Vo a=-g and s= hmax

=> we get hmax= Vo^2/ 2*g

using Vo=1/2*g*t from above eqns.. we get hmax= [1/2*g*t]^2 / 2*g = 1/8*g*t^2

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