Question

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.15 s later. Ignore air resistance.
a) If the height of the building is 19.6 m, what must the initial speed of the first ball be if both are to hit the ground at the same time
b) what must the height of the building for both balls to reach the groung at the same time if the initial velocity of the first ball is now 8.6 m/s​

Answer 1

Answer:

a) v₀ = 9.2 m/s

b) y₀ = 7.9 m

Explanation:

The position of the balls is given by the equation:

$y =- \frac{1}{2} gt^2 + v_0 t + y_0$

where:

acceleration g = 9.8 m/s²

time t

initial velocity v₀

initial height y₀

a) lets divide (a) in two parts:

1.part: How long will it take the second ball to fall down?

$v_0 = 0, y = 0\\0=- \frac{1}{2} gt^2 + y_0\\ t = \sqrt{\frac{2y_0}{g}}$

2. part: At time t from part1 + 1.15s, the first ball should land on the ground.

$y = 0, y_0 = 19.6, t = \sqrt{\frac{2y_0}{g}} + 1.15\\ 0 =- \frac{1}{2} gt^2 + v_0t + y_0$

This leaves only one unknown: v₀

$v_0 =\frac{1}{t}(\frac{1}{2} gt^2 - y_0)\\ v_0 = 9.2 \frac{m}{s}$

b)again, lets divide in two parts

1.part: Where will ball1 be relative to ball2 in 1.15s:

$t = 1.15s, v_0 = 8.6 m/s\\y= -\frac{1}{2} gt^2 + v_0t + y_0\\ \delta y = y - y_0 =v_0t -\frac{1}{2} gt^2$

and how fast will it go:

$v' = -gt + v_0$

2.part: Now we can plug in to the equation for the position of the two balls. Let's start with the second ball first:

$0 = -\frac{1}{2} gt^2 + y_0\\ y_0 = \frac{1}{2} gt^2$

Now let's use this result in the equation for the first ball:

$0 = - \frac{1}{2} gt^2 + v't + y_0 + \delta y = - \frac{1}{2} gt^2 + v't + \frac{1}{2} gt^2 + \delta y\\ 0 = v't + \delta y\\ t =- \frac{\delta y}{v'} \\ y_0 = \frac{1}{2} g(\frac{\delta y}{v'})^2\\ y_0 = 7.9m$