Question

HELP + extra pts // Two 10-m high diving platforms are at opposing ends of a 30-m pool. How fast must two clowns run straight off their boards if they want to collide at the middle of the pool and the surface of the water?

Answer 1

Explanation:

The clowns need to leave the diving boards with enough horizontal velocity such that each travels 15 m (half the width of the pool) in the same time that they fall (vertically) the 10-m from the top of the diving board.

We'll assume no force acts on the clowns horizontally to slow them down while they are in flight. And we'll assume that only gravity acts on the clowns vertically.

We can treat the horizontal and vertical components separately. This will help simplify the problem.

Let's start with the vertical displacement. Let's say the clown is dropped from a height of 10-m. How long would it take them to fall that distance?

Using our equations of motion (with constant, linear acceleration), we can solve for this time. $d = vt + \frac{1}{2} at^2$

Where d is the distance travelled, v is the initial velocity, a is acceleration, and t is time.

If the clown is dropped, they have an initial velocity of 0 (zero). We assumed only gravity acts on the clown, so acceleration equals the gravitational acceleration on earth. $a = 9.8m/s^2$

The distance the clown travels is 10-m from the diving board to the surface of the water.

Let's solve. $10 = 0*t + \frac{1}{2} (9.8) t^2 \rightarrow 20/9.8 = t^2 \rightarrow t = \sqrt{20/9.8}$

Now that we know time, we can calculate how fast the clown needs to be running when they leap from the diving board to cover a distance of 15-m (Remember, half the width of the pool.)

Using our equations of motion, we know that $d = vt + 0.5at^2$

We assumed no forces act horizontally on the clown, therefore a = 0. We just need to solve for v. Substituting in the time we just solved for, we get something like this. $15 = v \sqrt{20/9.8}$

I'll leave it to you to solve this equation.