Question

Se lanza una pelota y regresa al punto de partida 2.5 s después ¿Que altura máxima alcanzara? ¿a que altura se hallaba 1 segundo después del lanzamiento?

Answer 1

To determine the maximum height reached by a ball thrown vertically, we need to calculate its initial velocity and then use it to find the maximum height. Without the initial velocity provided, we can deduce it from the total time of flight. However, even with the information given in the scenarios, further detail is required to solve the problem accurately.

Step-by-step explanation:

Max Height and Height After 1 Second

If a ball returns to the same point after 2.5 seconds, the time taken to reach the maximum height is half of this time, which is 1.25 seconds. Using the formula for upward motion (h = vit - ½gt2) where g is the acceleration due to gravity (9.8 m/s²), and the initial velocity vi is zero at the maximum height, we can calculate:

hmax = 0 - ½(9.8 m/s²)(1.25 s)2
= -7.8125 m (which is not a valid answer as height cannot be negative. Instead, it indicates we'd need the initial vertical velocity to solve this correctly).

However, without the initial vertical velocity, we cannot find the maximum height. Besides, the negative sign indicates an incorrect approach as we only considered the descent part of the equation. Let's revise our process.

For a complete trajectory (without resistance and assuming a start from the ground level), the ball will spend half the total time ascending and half descending. Since it takes 2.5 s for the complete journey, it will spend 1.25 s to reach the maximum height. Using the fact that the final velocity at maximum height is 0, we can employ the kinematic equation (vf = vi + a*t) where vf is final velocity, a is acceleration, and t is time. Rearranging this, we find the initial velocity:

v

i

= v

f

- a*t
= 0 - (-9.8 m/s² * 1.25 s)
= 12.25 m/s.

The height after 1 second can now be found using the initial velocity and the acceleration due to gravity:

h1s = vit - ½gt2
= 12.25 m/s * 1s - ½(9.8 m/s²)(1s)2
= 7.35 m.

Answer 2

a)
`y_(max) = 7.7 m`

b) y = 7.4 m

Step-by-step explanation:

a) La altura máxima que alcanza la pelota se puede encontrar usando la siguiente ecuación:

`y_(max) = y_(0) + v_(0y)t - (1)/(2)gt^(2)` (1)

En donde:

`y_(0)`: es la altura inicial = 0

`v_(0y)`: es la velocidad inicial en y

g: es la gravedad = 9.81 m/s²

t: es el tiempo

Primero debemos encontrar la velocidad inicial en y:

`v_(fy) = v_(0y) - gt`

Sabiendo que la velocidad final en y es igual a cero (en la altura máxima), y con t = 1.25 s (la mitad del tiempo total) tenemos:

`v_(0y) = gt = 9.81 m/s^(2)*1.25 s = 12.3 m/s`

Ahora, podemos encontrar la altura máxima:

`y_(max) = v_(0y)t - (1)/(2)gt^(2) = 12.3 m/s*1.25 s - (1)/(2)9.81 m/s^(2)*(1.25 s)^(2) = 7.7 m`

b) Podemos encontrar la altura cuando t = 1 s con la misma ecuación (1):

`y_(max) = y_(0) + v_(0y)t - (1)/(2)gt^(2) = 12.3 m/s*1 s - (1)/(2)9.81 m/s^(2)*(1 s)^(2) = 7.4 m`

Por lo tanto, la pelota se encontraba a 7.4 m luego de 1 segundo despúes del lanzamiento.

Espero que te sea de utilidad!