Question

Mary needs to throw a can straight up into the air and then hit it with a second can. She needs the collision to happen at a height $h=5.0m$ above the point of throw. In addition, she is knowing that she requires ${{t}_{1}}=4s$ between the successive throws. Let us assume that she is throwing both the cans with identical speed. The acceleration due to gravity be $g=9.8m{{s}^{-2}}$
A) How long will it take to collide (in$s$) after the first can has been thrown into the air for the two cans? (Find your answer up to four significant figures)
B) Calculate the initial speed of the cans (in $m{{s}^{-1}}$)

Answer 1

The newton’s equations of motion has to be used in this question in order to solve it. The first equation of motion says that the final velocity will be equivalent to the sum of the initial velocity and the product of the acceleration and time taken. This may help you in answering this question.

Complete step by step solution:
Let the height of the first can be $x$, that of the second can be $y$ and both cans be thrown at speed $v$. Therefore we can write that,
$x\left( t \right)=v\left( t \right)-g\dfrac{{{t}^{2}}}{2}$
And also we can write that,
$y\left( t \right)=v\left( t-{{t}_{1}} \right)-g\dfrac{{{\left( t-{{t}_{1}} \right)}^{2}}}{2}$
As we all know, as per mentioned in the question,
$x\left( t \right)=y\left( t \right)=h$
From the first equation of height we can write that,
$v=\dfrac{h}{t}+\dfrac{gt}{2}$
Let us substitute this in the second equation of the height given as,
$h=\left( \dfrac{h}{t}+\dfrac{gt}{2} \right)\left( t-{{t}_{1}} \right)-\dfrac{g{{\left( t-{{t}_{1}} \right)}^{2}}}{2}$
The required height for the collision to take place can be shown as,
$h=5.0m$
The time taken can be written as,
${{t}_{1}}=4s$
The acceleration due to gravity has been mentioned as,
$g=9.8m{{s}^{-2}}$
Now let us substitute the values inside this equation which can be shown as,
$5=\left( \dfrac{5}{t}+\dfrac{9.8t}{2} \right)\left( t-4 \right)-\dfrac{9.8{{\left( t-4 \right)}^{2}}}{2}$
Simplifying this equation will give,
$t=2s$
B) As the cans should get collided at a height of $5m$,
$s=\dfrac{v\left( 4v-20 \right)}{40}-\dfrac{1}{2}\times 10\times {{\left( \dfrac{4v-20}{40} \right)}^{2}}$
Simplifying this equation can be written as,
$100={{v}^{2}}-5v-{{\left( v-5 \right)}^{2}}$
That is we can write that,
$100=5v-25$
Hence the initial speed of the cans will be,
$v=25m{{s}^{-1}}$

Note: The acceleration of a body can be found by taking the rate of the variation of the velocity with respect to time taken. The velocity can be found by taking the rate of the variation of the displacement with respect to the time taken. The displacement of the body is the perpendicular distance between the initial and final locations of the body. All these quantities are vector quantities possessing both magnitude as well as direction.