What Is the Initial Velocity of a Ball Thrown Upwards from a Building?

[petal5]

The question is as follows:

A man stands at the edge of the roof of a building 20.0m above the ground.He throws a ball vertically straight upwards(it just misses the edge of the building as it falls back down).If the total time taken for the ball to travel from his hand to the ground is 5.5seconds what is the initial velocity of the ball?

My question is,is it correct to use the following formula to solve the problem: x=xo + vot + 1/2at^2

Thanks!

 

[gunblaze]

Yes, it is. However, do be careful when dealing with the directions. Define ur +ve and -ve directions first.

 

[kyle8921]

I would say that the formula you have mentioned, x=xo + vot + 1/2at^2, is indeed applicable to solve this problem. This formula represents the displacement of an object in motion, where x is the final displacement, xo is the initial displacement, vo is the initial velocity, t is the time, and a is the acceleration. In this case, the acceleration due to gravity can be assumed to be a constant -9.8m/s^2.

To solve for the initial velocity, we can rearrange the formula as vo = (x-xo)/t - 1/2at^2. Substituting the given values, we get vo = (20-0)/5.5 - 1/2(-9.8)(5.5)^2 = 3.64m/s.

Therefore, the initial velocity of the ball is 3.64m/s. However, it is important to note that this is the vertical velocity of the ball. If we want to calculate the total initial velocity, we will need to consider the horizontal component as well. This would require more information about the angle at which the ball was thrown and the horizontal distance it traveled.

Overall, using the given formula is a valid approach to solving this problem, but it may not give us the complete picture of the ball's motion. As scientists, we must always consider all factors and variables in our calculations to accurately describe and understand the phenomena we are studying.