Question

A stone is dropped by a person from the top of a building which is 200m tall at the same time another stone was thrown upwards with a velocity of 50m/s by a person standing at the foot of the building find the time the stones meet

Answer 1

$\text{Let n be a function showing the distance from}\newline \text{ ground level to stone.}$

$\text{For a stone thrown from a building:}$

$h_1(t)= 200-\frac{gt^2}{2};$

$\text{for a stone thrown from the ground:}$

$h_2(t)=V_0t-\frac{gt^2}{2}= 50t -\frac{gt^2}{2}$

$\text{When stones meet:}$

$h_1(t) = h_2(t)$

$200-\frac{gt^2}{2}=50t-\frac{gt^2}{2};$

$t=4$

Answer:4 seconds meeting time of stones