Answer to Question #181256 in Physics for John Aron Calag Orozco

Question

A balloon rises from the ground with the constant acceleration of 4ft per sec^2. five seconds, later, a stone is thrown vertically up from the launching site. what must be the minimum initial velocity of the stone for it to just touch the balloon? Note that the balloon and stone have the same velocity and contact.

Answer 1

The balloon moves according to the following equation:

$$h(t)=\frac{at^2}{2}.$$

The stone thrown with initial velocity $v$ moves according to a similar equation:

$$h(T)=vT-\frac{gT^2}{2}.$$

If we take the fact that it was launched 5 s later into account $(T=t-5)$, we can re-write the equation as

$$h(t)=v(t-5)-\frac{g(t-5)^2}{2}.$$

Equate:

$$\frac{at^2}{2}=v(t-5)-\frac{g(t-5)^2}{2}$$

So, we have an equation with two unknowns: $v, t$. Introduce another equation that describes the fact that their velocities at the impact must be equal:

$$v_b=at,\\ v_s=v-gt,\\ v_b=v_s,\\ at=v-gt,\\ v=t(a+g).$$

Substitute:

$$\frac{at^2}{2}=t(a+g)(t-5)-\frac{g(t-5)^2}{2}.$$

Solve for time:

$$t=5.3\text{ s}.$$

Minimum initial velocity of the stone:

$$v=5.3(4+32.174)=191.7\text{ ft/s}.$$