Mathematics, 29.06.2019 00:30 ellemarshall13

Round answer to the nearest tenth. a hang glider dropped his cell phone from a height of 350 feet. how many seconds did it take for the cell phone to reach the ground?

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Answer from: nadia74

The cell phone takes about 4.7 seconds to reach the ground

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

Height = h = 350 feet

Gravitational Acceleration = g = 32.2 feet/s²

Initial Velocity = u = 0 m/s

Unknown:

time taken to reach the ground = t = ?

Solution:

This problem is about Free Fall.

When an object experiences free fall, the acceleration experienced by the object is the same as the acceleration due to gravity which is equal to 9.8 m/s² or 32.2 feet/s²

$h = ut + \frac{1}{2}gt^2$

$350 = 0t + \frac{1}{2}(32.2)t^2$

350 = 0 + 16.1t^2

$t^2 = 350 \div 16.1$

$t \approx 4.7 ~ \text{seconds}$

Learn moreVelocity of Runner : Kinetic Energy : Acceleration : The Speed of Car : Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

Round answer to the nearest tenth. a hang glider dropped his cell phone from a height of 350 feet. h

Answer from: Quest

The answer is the first one

Answer from: Quest

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf =

pf = 6,25 u

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step-by-step explanation:

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