Question

Una piedra es lanzada horizontalmente desde una ventana con una velocidad inicial cuya magnitud es de 10m/s y cae al suelo despues de 5 segundos
a que altura se encuentra la ventana?
a que distancia cae la piedra de la base del edificio?

Answer 1

Answers:

a) 122.5 m

b) 50 m

Explanation:

Here we are dealing with projectile motion, where we can use the following two equations that model the parabolic motion of the stone after being thrown from the window:

$h=h_{o}+V_{o}sin(\theta)t-\frac{1}{2}gt^{2}$ (1)

$x=V_{o}cos(\theta)t$ (2)

Where:

$h=0 m$ is the height of the stone when it hits the ground

$h_{o}$ is the initial height of the stone

$V_{o}=10 m/s$ is the initial velocity of the stone

$\theta=0\°$ since we are told the stone was thrown horizontally

$g=9.8 m/s^{2}$

$t=5 s$ is the time the stone is in air

$x$ is the horizontal distance the stone travelled

Knowing this, we can answer the questions:

a)Initial height of the stone:

In this case we will isolate $h_{o}$ from (1):

$h_{o}=\frac{1}{2}gt^{2}$ (3)

$h_{o}=\frac{1}{2}(9.8 m/s^{2})(5 s)^{2}$ (4)

$h_{o}=122.5 m$ (5) This is the initial height

b) Horizontal distance:

In this case we will use equation (2):

$x=(10 m/s)cos(0\°)(5 s)$ (6)

$x=50 m$ (7) This is the horizontal distance