Se lanza una pelota hacia arriba a una velocidad de 15 m/s. Calcula: El tiempo que tardará en subir.La altura máxima alcanzada.El tiempo que empleará en bajar.

1. Break Down the Problem

We need to calculate three things:

1. The time it takes for the ball to rise to its maximum height.
2. The maximum height reached by the ball.
3. The time it takes for the ball to fall back to the original position.

2. Relevant Concepts

• Kinematic Equations: We will use the following kinematic equations for motion under uniform acceleration (gravity).
  • $v = u + at$
  • $s = ut + \frac{1}{2} a t^2$
• Where:
  • $v$ = final velocity (0 m/s at the top of the flight)
  • $u$ = initial velocity (15 m/s)
  • $a$ = acceleration (-9.81 m/s², as gravity acts downward)
  • $t$ = time
  • $s$ = displacement (maximum height in this case)

3. Analysis and Detail

A. Time to Reach Maximum Height

Using the equation $v = u + at$: $0 = 15 + (-9.81) t$ $9.81 t = 15$ $t = \frac{15}{9.81} \approx 1.53 \text{ seconds}$

B. Maximum Height

Using the equation $s = ut + \frac{1}{2} a t^2$: $s = 15(1.53) + \frac{1}{2} (-9.81)(1.53)^2$ Calculating each term: $s = 22.95 + \frac{1}{2} (-9.81)(2.3409)$ $s = 22.95 - 11.48 \approx 11.47 \text{ meters}$

C. Time to Fall Back

The time taken to fall back is the same as the time taken to rise. Thus, the time to fall is also approximately: $t = 1.53 \text{ seconds}$

4. Verify and Summarize

• Time to rise: $\approx 1.53$ seconds
• Maximum height: $\approx 11.47$ meters
• Time to fall: $\approx 1.53$ seconds

Final Answer

1. Time to rise: $\approx 1.53$ seconds
2. Maximum height: $\approx 11.47$ meters
3. Time to fall: $\approx 1.53$ seconds