Question

Mary wants to throw a can staight up into the air and then hit it with a second can she want the collusion to occur at height h=5.0m above the throw point . In addition she knows that she needs t1=4.0secs between successive throws . Assume that she throws both can with the same speed take g to be 9.81m/per seconds square
a. How long it takes (in seconds ) after the first can has been thrown into the air for the two cans to collide
B. Find the initial speed of the cans (in meters/ seconds )

Answer 1

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Answer 2

Answer: a) t = 4.240s , B) v(t=0) = 21.98m/s

Explanation: Let t be the time when they are in the same spot (colision), and t any other arbitrary time. x is the position function, so for example x(t=0) will be the initial position. v is the velocity function and a=-g

The equation for the position of the first can is

x(t)= x(t=0)+v(t=0)t+(a/2)t^2 solve for v(t=0)

v(t=0)= ( x(t)/ t ) - (at/2)

The equation for the position of the second one is

x(t)= x(t=0)+v(t=0)*(t-4)+(a/2)*(t-4)^2),

When t=t by definition we know they are at the same place, we don't know two variables. The initial speed on both objects its the same so we can substitude our v(t=0) in the equation of the second can.

x(t)= x(t=0)+ ( x(t)/ t ) - (at/2)*(t-4)+(a/2)*(t-4)^2

substitude with known values

5 = 0+ ( 5/ t ) - (-9.81 t/2)*(t-4)-9.81/2(t-4)^2

you get a quadratic out of that

39.24t^2-156.96t-40=0

When you solve it, you take the greatest value because this needs to be true t >4, then t = 4.24039 approximately

Now use the equation of the first can to get the initial velocity

substitude with known values

5 = 0 +v(t=0)(4.24039)+(-9.81/2)(4.24039)^2

v(t=0)= 21.97824

We round to 4 sigfigs and t = 4.240s , v(t=0) = 21.98m/s

If you want to verify that its correct, plug the two equations into a graph, you get two parabolas, the y axis is going to be position, the x axis is going to be time, in the point where they intersect with each other  (colide in this case) the y coordinate should be really close to 5 (the height), and the x coordinate should be the time t.

y=21.98x-4.905x^2 (first can)

y=21.98(x-4)-4.905(x-4)^2  (second can)

(to graph)