Question

Wile standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 6.79m/s. the stone is subsequently falls to the ground, which is 12.1m below the point of where the stone leaves your hand. How much time is the stone in the air?

Answer 1

Solve:

First the stone will move upward till it reaches a maximum height and after that, it will start to fall down and at last hit the ground. So, we can divide the problem into two. first, when it goes up and second when it comes down.

When the stone goes up, to calculate the time required to reach maximum height we can use the following equation:

v = u + a*t

Here, v = final velocity which will be zero at the maximum height.

u = initial velocity = 6.79 m/s.

a = -g (acceleration due to gravity) = -9.81 m/s²

Substituting this in the above equation we get,

0 = 6.79 - 9.81*t

t = 6.79/9.81

t = 0.692 seconds.

Now, we will find how much distance it traveled during this time and this can be found using:

d = u*t + 0.5*a*t²

Substituting the values we get,

d = 6.79*0.692 - 0.5*9.81*0.692²

d = 2.35 meters.

So, after reaching this height, the stone will start falling and cover a total distance of (H = 2.35 + 12.1 = 14.45) meters before reaching the ground.

When the stone is falling down, for this case, we have, u = 0 and a = g = 9.81.

Then using the equation:

H = u*t + 0.5*a*t²

14.45 = 0 + 0.5*9.81*t²

we get,

t = 1.716 seconds.

So, total time = 0.692 + 1.716 = 2.408 seconds.

Thank you.