Stone is thrown from the top of the building.

[patelneel1994]

Homework Statement

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

Vi = -20 m/s height = 60 m
t = ?

Homework Equations

d = Vi.t + 1/2 a t2

The Attempt at a Solution

Couldn't figure out. How to use height as a distance?

 

[BvU]

It has the same dimension ! The equations are OK.

 

[patelneel1994]

So 60 m = -20m/s . t + 1/2 a t2
60 m = -20m/s t - 4.9 m/s2 t2?
What should I do after this ...

 

[collinsmark]

So 60 m = -20m/s . t + 1/2 a t2
60 m = -20m/s t - 4.9 m/s2 t2?

You seem to be using the convention that up is positive, which is fine. But is the stone traveling 60 m downward or upward? What does that tell you about the sign of the 60 m in your equation?

What should I do after this ...

Solve for t. (Hint: quadratic equation.)

 

[HalfLostAndSearching]

I would approach this problem by first setting up a coordinate system with the ground as the origin and the positive direction pointing upwards. This will allow us to use the given initial velocity and height as distances in our calculations.

Next, we can use the equation d = Vi.t + 1/2 a t^2 to solve for time. We know that the initial velocity, Vi, is -20 m/s (since it is pointed downward), and the acceleration, a, is equal to the acceleration due to gravity, which we can assume to be -9.8 m/s^2. We also know that the final distance, d, is equal to the height of the building, 60 m.

So we can plug these values into the equation and solve for time:

60 m = (-20 m/s) t + 1/2 (-9.8 m/s^2) t^2

Simplifying, we get t^2 - 4t + 12 = 0.

Using the quadratic formula, we can find two possible solutions for t: t = 4 seconds or t = 3 seconds. However, we can disregard the negative solution (t = 3 seconds) since time cannot be negative in this case. Therefore, the time elapsed between the instant of release and the instant of impact with the ground is approximately 4 seconds.

In conclusion, as a scientist, I would approach this problem by setting up a coordinate system, using the equation d = Vi.t + 1/2 a t^2, and solving for time. By doing so, we can determine that the time elapsed between the instant of release and the instant of impact with the ground is approximately 4 seconds.