# Thread: Initial speed and time

1. ## Initial speed and time

OK, here's the problem: A ball (first ball) is dropped from rest from a height 20.0m above the ground. Another ball is thrown (second ball) vertically upward from the ground at the instant the first ball is released. Determine the intial speed of the second ball if the two balls are to meet at a height 10.0m above the ground.

The equations I am using are time = square root(2y/-g), and initial velocity = x/t.

For time I get square root((2*(-10m))/9.8m/s^2) = 1.43 seconds.
For initial velocity of the second ball I get 10m/1.43s = 6.99m/s.

However, at the end of the problem it says that the answer should be 14m/s but I can't figure out how to get that so am I doing something wrong, or is the answer I am given wrong?

2. you cannot apply $\displaystyle S=ut$ when there is acceleration. So to find initial velocity, apply $\displaystyle S=ut+\frac{1}{2}at^2$

3. Originally Posted by learningguy
OK, here's the problem: A ball (first ball) is dropped from rest from a height 20.0m above the ground. Another ball is thrown (second ball) vertically upward from the ground at the instant the first ball is released. Determine the intial speed of the second ball if the two balls are to meet at a height 10.0m above the ground.

The equations I am using are time = square root(2y/-g), and initial velocity = x/t.

For time I get square root((2*(-10m))/9.8m/s^2) = 1.43 seconds.
For initial velocity of the second ball I get 10m/1.43s = 6.99m/s.

However, at the end of the problem it says that the answer should be 14m/s but I can't figure out how to get that so am I doing something wrong, or is the answer I am given wrong?

$\displaystyle y_1(t)=20-\frac{g}{2}t^2$ and
$\displaystyle y_2(t)=v_0t-\frac{g}{2}t^2$

Using the first equation gives

$\displaystyle \displaystyle 10=20-\frac{g}{2}t^2 \iff t=\pm 2\sqrt{\frac{5}{g}}$

putting this into the 2nd equation gives

$\displaystyle \displaystyle 10=v_0\left( 2\sqrt{\frac{5}{g}}\right)-\frac{g}{2}\left( 2\sqrt{\frac{5}{g}}\right)^2 \implies v_0=2\sqrt{5g}$

4. Originally Posted by learningguy
OK, here's the problem: A ball (first ball) is dropped from rest from a height 20.0m above the ground. Another ball is thrown (second ball) vertically upward from the ground at the instant the first ball is released. Determine the intial speed of the second ball if the two balls are to meet at a height 10.0m above the ground.

The equations I am using are time = square root(2y/-g), and initial velocity = x/t.

For time I get square root((2*(-10m))/9.8m/s^2) = 1.43 seconds.
For initial velocity of the second ball I get 10m/1.43s = 6.99m/s.

However, at the end of the problem it says that the answer should be 14m/s but I can't figure out how to get that so am I doing something wrong, or is the answer I am given wrong?
Your time is correct. However, your other equation is wrong because there is a constant, acceleration due to gravity, in place (assuming it is on Earth). As a result you have to use the Δy=viΔt + 1/2a(Δt)^2 formula. Then you have to re-arrange it to solve for vi. Remember, to use your vectors correctly.