A ball is dropped from a height of 200m and one second later another ball is thrown vertically upwards with a speed of 50 m/s. At what height from the ground and after how many seconds of the release of the first ball will the two balls cross each other.

Question

A ball is dropped from a height of 200m and one second later another  ball is thrown vertically upwards with a speed of 50 m/s. At what  height from the ground and after how many seconds of the release of  the first ball will the two balls cross each other.

anonymous · Answer

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current height of the first object can be described with equation:
h = h0 - (g*t^2 / 2)      -> h0 is initial height (200m), and (g*t^2 / 2) is gravitational force    forcing the object down to the ground 

current height of the second object can be described with equation:
h = v0*t - (g*t^2 / 2)      -> v0 is initial speed of the second object (50m/s), and again (g*t^2 / 2) is gravitational force forcing the object in the opposite direction, down to the ground 

we want to know when will the heights be the same, so we equal the two equations, and we have to change the t in second equation with t-1, because the second object is launched one second later

h0 - (g*t^2 / 2) = v0*(t-1) - (g*(t-1)^2 / 2) 
200 - (9.81*t^2 / 2) = 50 * (t-1) - (9.81*(t-1)^2 / 2)
200 - 4.905 * t^2 = 50t - 50 - 4.905 * (t^2 - 2t + 1)
200 - 4.905 * t^2 = 50t - 50 - 4.905 * t^2 + 9.81t - 4.905
200 = 50t - 50 + 9.81t + 4.905
254.905 = 59.81t
t = 4.262 sec

we will get the height by inserting the time in one of the equations:

h = h0 - (g*t^2 / 2) = 200 - (9.81*(4.262^2) / 2) = 110.9 m