Freefall problem with object shot up off a building.

[VinnyCee]

A rock is shot vertically up froma building and starts falling back to Earth after 1.60 seconds. After 6.00 seconds, it hits the ground. Constant acelleration of 9.81 m / s^2.

a. What is initial velocity?

b. What is max height above building?

c. How tall is the building?

I got the following answer, I am unsure of them though:

a => 15.7 m / s
b => 12.6 m
c => 164 m

Could anyone tell me if my figures are correct?
Thank you.

 

[Integral]

a & b look good. c not so. Please show me your work.

 

[M98Ranger]

Your figures are close, but not entirely correct. Here is a breakdown of the correct calculations for each part:

a. To find the initial velocity (Vo), we need to use the formula Vf = Vo + at, where Vf is the final velocity (which in this case is 0 m/s), a is the acceleration (9.81 m/s^2), and t is the time (1.60 seconds).

So, plugging in the values, we get: 0 = Vo + (9.81)(1.60). Solving for Vo, we get: Vo = -15.7 m/s.

The negative sign indicates that the initial velocity is in the opposite direction of the acceleration, which makes sense since the rock is shot vertically upwards.

b. To find the maximum height (h) above the building, we can use the formula h = (Vo^2)/(2a). Plugging in the values, we get: h = (-15.7)^2 / (2 * -9.81) = 12.6 m.

c. To find the height of the building, we can use the formula h = (Vo^2)/(2a) + (Vo)(t), where t is the total time (6.00 seconds). Plugging in the values, we get: h = (-15.7)^2 / (2 * -9.81) + (-15.7)(6.00) = 164 m.

So, your calculations for b and c are correct, but for a, the initial velocity should be -15.7 m/s. This is because the initial velocity is in the opposite direction of the acceleration, so it has a negative value.

I hope this helps clarify the problem for you!