Mary wants to throw a can straight up into the air and then hit it with a second can. She wants the collision to occur at height h=5.0 m above the throw point. In addition, s/he knows that she needs t1=3.0 s between successive throws. Assuming that s/he throws both cans with the same speed. Take g to be 9.81 m/s2.
hint: you might want to look at Professor's Lewin's Help Session before you attempt this problem.
(a) How long it takes (in s) after the first can has been thrown into the air for the two cans to collide?
Δt=
(b) Find the initial speed of the cans (in m/s).
v0=
4.981
Please help with this question!!
trfhyd
To solve this problem, we need to analyze the motion of the two cans separately and find the conditions that will lead to a collision at height h = 5.0 m.
Let's start with the first can, which is thrown straight up. We can use the equation of motion:
h = v0 * t - 0.5 * g * t^2
where h is the height, v0 is the initial velocity, t is the time, and g is the acceleration due to gravity.
After solving this equation for t, we get:
t = (v0 / g) + sqrt((v0^2 / g^2) + 2h / g)
Now let's consider the second can. We need the time it takes for the second can to hit the same height h = 5.0 m after the first can is thrown. Since the time between successive throws is t1 = 3.0 s, the second can will be thrown t1 seconds after the first can. Therefore, the time for the second can to reach height h is:
t2 = t1 + t
Now we have the time for both cans, and we need to find the conditions for a collision. For a collision to occur, the second can must be at the same height as the first can, so we set h equal to 0 in the equation of motion for the second can:
0 = v0 * t2 - 0.5 * g * t2^2
Simplifying this equation, we get:
0.5 * g * t2^2 - v0 * t2 = 0
This is a quadratic equation in t2. We can solve it using the quadratic formula:
t2 = (-(-v0) +/- sqrt((-v0)^2 - 4 * 0.5 * g * 0)) / (2 * 0.5 * g)
Now we can substitute t2 = t1 + t and solve for t:
t1 + t = (-(-v0) +/- sqrt((-v0)^2 - 4 * 0.5 * g * 0)) / (2 * 0.5 * g)
Simplifying further, we get:
t = (v0 +/- sqrt(v0^2 - 2 * g * t1)) / g
Now we can solve part (a):
(a) How long does it take (in s) after the first can has been thrown into the air for the two cans to collide?
The time after the first can has been thrown for the two cans to collide is given by:
t = (v0 +/- sqrt(v0^2 - 2 * g * t1)) / g
(b) Find the initial speed of the cans (in m/s).
To find the initial speed of the cans, we substitute t into the equation:
v0 = g * t +/- sqrt(v0^2 - 2 * g * t1)