# Can you help me solve a difficult kinematics problem? [closed]

I've been working on this problem for a while now, but I still don't know how to solve it. It involves 2D kinematics, which aren't in themselves hard, but this one threw me for a loop.

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 seconds later (ignore air resistance).

(a) If the height of the building is 20.00 meters, what must be the initial speed of the first ball if both are to hit the ground at the same time?

(b) Consider the same situation, but now let the initial speed of the first ball be given and treat the height of the of the building as an unknown. What would the height of the height of the building for both balls to reach the ground at the same time?

Now I know for (a) the initial position of each ball, the acceleration and velocity of the second ball (gravity), and final position of each ball, but that's it! I have no idea how to find out the velocity of the first ball. I'm pretty sure that you have to calculate the time it takes for the first ball to reach 20 meters in a second, to be dropped at the same time as the first ball, but I have no idea.

And in (b), the initial speed isn't given! How do I solve that?!

• I've added the homework tag. In the future, please use that tag on questions of this type. – user4552 Sep 11 '13 at 5:09

All you need for solve this problem is the equation for an uniformly accelerated motion: $$x(t)=\frac{1}{2}at^2+v_0\cdot t+x_0$$ where the motion takes place on a straight line if the acceleration and the speed are collinear (i.e. on the same direction). $x_0$ and $v_0$ are the initial position (the roof) and the initial speed. Let me add different subscript (1 and 2) for the two balls: $$\begin{cases}x_1(t_1)=\tfrac{1}{2}at_1^2+v_{1(0)}t_1+x_{1(0)}\\ x_2(t_2)=\tfrac{1}{2}at_2^2+x_{2(0)}\end{cases}.$$ $a=-g=-9.81...$ m/s$^2$, since the gravitational acceleration points down (I'm thinking that 'up' is positive and 'down' is negative, as a custom choice), $v_{2(0)}=0$, hence I will omit any subscript on the speed of the first ball, $v_{1(0)}=v_0$; $x_{1(0)}=x_{2(0)}=x_0=$ the eight of the roof, and $t_2=t_1-1$ sec. (since the second ball starts one second later than ball 1, i.e. it has one second less) then I simply call $t$ the time of the motion of the first ball and substitute $t_2=t-1$ sec. in the previous equation: $$\begin{cases}x_1(t)=-\tfrac{1}{2}gt^2+v_0t+x_0\\ x_2(t)=-\tfrac{1}{2}g(t-1)^2+x_0\end{cases}.$$ You ask that a certain time $t$ your balls are in the same place, that is: $x_1(t)=x_2(t)$, or, more precisely, they are on the ground (which we can assume is at zero eight): $x_{1,2}(t)=0$. Then you have to solve this equations: $$\begin{cases}0=-\tfrac{1}{2}gt^2+v_0t+x_0\\ 0=-\tfrac{1}{2}g(t-1)^2+x_0\end{cases}$$ where both $t$ and $v_0$ are unknown. Some maths is all you need to solve this problem!

SOLUTIONS:

a) $v_0=\frac{1}{2}g\left(\sqrt{2x_0/g}+1\right)-x_0/\left(\sqrt{2x_0/g}+1\right)\simeq 8.17...$ m/s

b) You have to solve the previous equation for $x_0$...

$x_0=\left[\left({1\over2}g-v_0\right)/\left(\sqrt{2/g}-\sqrt{2\cdot g}\right)\right]$

Answer for a: The second ball is released from rest and assumes going down is negative: u=0 g=-9.81 v=? h=-20 t=? By using the kinematics equation substitute the value in following equation: $$s=ut^2+\frac12 at^2$$ So: $$-20=-\frac12 9.81 t^2$$ Therefore: t=2.02 s.

It said it released after 1 s so: t=3.02 s.

The first ball is projected vertically upward: u=? g=-9.81 h=-20 t=3.02.

Substitute the values: $$-20=u (3.02) - \frac12(9.81)(3.02)^2$$ Therefore: u=8.19 m/s.

• Don't answer homework calculation questions. – Brandon Enright Sep 30 '14 at 15:57