Solving for Building Height and Time in Horizontal Projectile Motion

[harujina]

Homework Statement

A ball is thrown horizontally from the top of a building and lands 20.0m away from it. If the ball is initially thrown at a velocity of 10.0m/s, how high is the building? How long does it take for the ball to reach the ground?

Homework Equations

dv = Viv * t + 1/2a * t

The Attempt at a Solution

dh (horiz.distance) = 20.0m
Vh = 10.0m/s
Viv (vertical init. velocity) = 0m/s
a = -9.8m/s
dv (vert. distance) = ?
t = ?

I'm confused because I'm missing two of the variables which are needed in the equation that will solve this problem...? (dv and t)

 

[NascentOxygen]

The horizontal velocity of the ball is fixed and constant during the whole time of flight. Knowing this, you can determine the time of flight.

 

[harujina]

Using the same equation as the one stated above? Because that one's giving me a totally wrong answer.

 

[haruspex]

Using the same equation as the one stated above? Because that one's giving me a totally wrong answer.

Yes, the same equation, but of course in the horizontal direction a = 0. pls post your working.

 

[Nickie]

I would first clarify the problem by stating that this is a horizontal projectile motion problem where the ball is thrown horizontally from the top of a building. I would also mention that the only force acting on the ball is gravity, which causes it to accelerate downwards at a rate of 9.8m/s^2.

To solve for the building height, we can use the equation dh = Vh * t, where dh is the horizontal distance, Vh is the initial horizontal velocity, and t is the time. In this case, we know dh and Vh, so we can rearrange the equation to solve for t. This gives us t = dh / Vh = 20.0m / 10.0m/s = 2.0s.

Now that we have the time, we can use the equation dv = Viv * t + 1/2a * t^2 to solve for the vertical distance. Here, dv is the vertical distance, Viv is the initial vertical velocity (which is 0 in this case), a is the acceleration due to gravity, and t is the time we just calculated. Plugging in the known values, we get dv = 0 * 2.0s + 1/2 (-9.8m/s^2) * (2.0s)^2 = -19.6m.

Since the vertical distance represents the height of the building, we can conclude that the building is 19.6m tall. To find the time it takes for the ball to reach the ground, we can use the equation dv = 1/2a * t^2 and solve for t. This gives us t = √(2 * dv / a) = √(2 * 19.6m / 9.8m/s^2) = √(4s^2) = 2.0s.

Therefore, it takes the ball 2.0 seconds to reach the ground. In summary, by using the equations for horizontal and vertical motion, we can solve for the building height and time in this horizontal projectile motion problem.