In how many seconds after being thrown does the rock strike the ground?

Question

A rock is thrown vertically upward with a speed of 14.0 m/s from the roof of a building that is 40.0 m above the ground. Assume free fall.

Answer 1

dx=v*t+1/2*a*t^2
x=distance [m]
v=velocity [m/s]
a=acceleration [m/s^2] = 9,8
t=time [s]

1)
dx=0=14*t+4.8*t^2
14=4.8t t=14/4.8= 2.92s


2)
dx=40=14*t+1/2*9.8*t^2
40=14*t+4.8*t^2
t=1.75s


1)+2)
2.92+1.75= 4.67s!!

Answer 2

Use the formula S = Ut + ½ a t^2.

Let us take the upward direction as positive.

Then S= - 40m (since it is downward)

a = - 9.8 m/s^2 (since the velocity decreases)

U = 14 m/s (since it is upward)

Therefore, -40 = 14 t - 0.5 x 9.8 x t^2

Or 4.9 t^2 - 14 t -40 = 0.

b^2 - 4 a c = 980

{-b + or - (b^2 - 4 a c) ^0.5} / 9.8 = 4.6229543 s and - 1.7658114 s.


The roots of the equation are 4.6229543 s and - 1.7658114 s.
The negative sign indicates that the object was on the ground before the starting time and the positive sign shows that the object will on the ground after start.

Answer 3

a hint, gravity 9.8 m/sec speed/velocity = distance divided by time , try hard.

Answer 4

Um this requires thinking... and i'm too lazy : ]