. Desde el borde de una azotea de un edificio, se lanza un cuerpo hacia abajo con una velocidad de 20 m/s, si el edificio mide 105 m. ¿Cuántos segundos dura la caída?

t = 3.01 s

Explanation:

In order to calculate how long it takes to the object to fall to the ground, you use the following formula, for the calculation of the height:

   (1)

yo: height of the building = 105 m

vo: initial velocity of the body = 20m/s

g: gravitational acceleration = 9.8m/s^2

t: time = ?

To find the time t, you take into account that when the body arrives to the ground the height is zero, that is, y = 0.

You replace the values of all parameters in the equation (1), and you obtain a quadratic polynomial for t:

Next, you use the quadratic formula to get the roots of the polynomial:

a = -4.9

b = -20

c = 105

You choose the positive value t1, because it has physical meanning.

Hence, the body takes 3.01s to arrive to the ground

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft