Answers:
(a) [tex]y(t)=300m-4.9m/s^{2}t^{2}[/tex]
(b) [tex]t=7.82s[/tex]
(c) [tex]V=-76.7 m/s[/tex]
(d) [tex]t=7.6s[/tex]
Explanation:
This problem is a good example of vertical motion, and the main equations for this situation are as follows:
[tex]y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (1)
[tex]V=V_{o}-gt[/tex] (2)
Where:
[tex]y[/tex] is the height of the stone at a given time
[tex]y_{o}=300m[/tex] is the initial height of the stone
[tex]V_{o}[/tex] is the initial velocity of the stone
[tex]t[/tex] is the time
[tex]g=9.8m/s^{2}[/tex] is the acceleration due to gravity on Earth
Now, for parts (a), (b) and (c) we are specifically talking about free fall (where the main condition is that the initial velocity must be zero [tex]V_{o}=0[/tex]).
How do we know this?
Because we are told "the stone is dropped".
Having this clear, let's start with the calculations:
In order to approach this, we will use equation (1), remembering the condition [tex]V_{o}=0[/tex]:
[tex]y(t)=y_{o}-\frac{1}{2}gt^{2}[/tex] (3)
[tex]y(t)=300m-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex] (4)
[tex]y(t)=300m-4.9m/s^{2}t^{2}[/tex] (5) This is the distance of the stone above ground level at time t
In this case, [tex]y=0[/tex], because we have to find the time when the stone finally hits the ground.
Rewritting (1) with this condition:
[tex]0=y_{o}-\frac{1}{2}gt^{2}[/tex] (6)
Isolating [tex]t[/tex]:
[tex]t=\sqrt{\frac{2y_{o}}{g}}[/tex] (7)
[tex]t=\sqrt{\frac{2(300m)}{9.8m/s^{2}}}[/tex] (8)
Then:
[tex]t=7.82s[/tex] (9)
In this part, rewritting equation (2) will be usefull, remembering [tex]V_{o}=0[/tex] and using the timewe found in (9), which is the time when the stone strikes the ground:
[tex]V=-gt[/tex] (10)
[tex]V=-(9.8m/s^{2})(7.82s)[/tex] (11)
[tex]V=-76.68m/s \approx -76.7 m/s[/tex] (12) Note the velocity has a negative sign because its direction is downwards.
Now the initial velocity is [tex]V_{o}=-2m/s[/tex] (downwards). So, let's use equation (1) again an find [tex]t[/tex]:
[tex]0=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (13)
[tex]0=300m+(-2m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}[/tex]
[tex]0=-4.9m/s^{2}t^{2}+(-2m/s)t+300m[/tex] (14)
At this point we have a quadratic equation of the form [tex]0=at^{2}+bt+c[/tex], and we have to use the quadratic formula if we want to find [tex]t[/tex]:
[tex]t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex] (15)
Where [tex]a=-4.9[/tex], [tex]b=-2[/tex], [tex]c=300[/tex]
Substituting the known values and choosing the positive result of the equation:
[tex]t=7.623 s \approx 7.6 s[/tex] This is the time it takes to the stone to reach the ground when [tex]V_{o}=-2m/s[/tex]
