Hi, new here, im really stuggling on how to even start this question any help would be much appreicated!

[jacksono]

Hi, new here, im really stuggling on how to even start this question any help would be much appreicated!​

A ball is thrown straight up from the edge of the roof of a building of height h, with an initial speed of v0. A second ball is dropped from the roof t0 later. Ignore air resistance. 2 (a) If the balls reach the ground at the same time, show that the height is

Thanks in advance!

 

[mmm4444bot]

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[HallsofIvy]

What DO you know about this problem? Do you know about gravity? That there is a downward force on every object proportional to its mass so that there is a constant downward acceleration of approximately g= 9.8 meters per second per second (or g= 32.2 feet per second per second).

The first ball is thrown up (from height h) with initial velocity \(\displaystyle v_0\) so, with acceleration g, after t seconds its speed will be \(\displaystyle v_0- gt\) and it will have reached height \(\displaystyle h+ v_0t- (g/2)t^2\). It will be 0 (the ball willl have hit the ground) when \(\displaystyle h+ v_0t- (g/2)t^2= 0\). That is a quadratic equation that has two roots, one positive and one negative. Since the ball cannot hit the ground until after it is thrown, solve for the positive root.

At time \(\displaystyle t_0\), a second ball is dropped from height h. It also has downward acceleration g but has 0 initiall speed so after time t has downard speed -gt and will be a height h- gt. That ball hits the ground when \(\displaystyle h- g(t-t_0)= 0\) or \(\displaystyle t= t_0+ h/g\).

"If the two balls hit the ground at the same time" set those times equal and solve for h.