A ball is thrown from the top edge of a building with initial velocity components of 15m/s(up) vertically and 20m/s horizontally. It strikes ground 140m from the base of the building. What is the height of the building?

Question

A ball is thrown from the top edge of a building with initial velocity components of 15m/s(up) vertically and 20m/s horizontally. It strikes ground 140m from the base of the building. What is the height of the building?

I tried using V2^2 = V1^2 + 2ad and it gave me 11.25 which I know is wrong. How would I calculate this?

Answer 1

since you have the horizontal or x velocity you can calculate the time needed for the ball to reach the x distance(140m)

using this equation:
xf-xi= v_i+ 1/2at^2

the thing is there is no acceleration in x direction so the a is 0 thus that part cancels out of the equation to make:xf-xi= v_i

Then use this equation, find t for the time.
After this, use t and then plug it into the equation again (same one) However now that you have t it took to reach the ground in x direction it is the (same in the y) direction
you can use the same equation but this time there IS a acceleration in the y direction.

yf-yi= v_i+ 1/2at^2

so use the same equation but now plug in initial v for the vertical or y direction and t that you found before and also plug in 9.8m/s^2 for the acceleration. And you should find your answer.

Answer 2

To solve this problem, you can use the equations of projectile motion. Let's break down the given information and step-by-step calculations to find the height of the building:

Given:
- Initial velocity component in the upward direction (Vy) = 15 m/s
- Initial velocity component in the horizontal direction (Vx) = 20 m/s
- Distance the ball strikes the ground (d) = 140 m

Step 1: Calculate the time taken for the ball to reach the ground.
To find the time of flight, we can use the vertical component of the velocity:
Vy = V0y + gt
0 = 15 − 9.8t (Since the ball is in free fall, the acceleration due to gravity is negative)

Solving for t, we get:
t ≈ 15/9.8 ≈ 1.53 seconds

Step 2: Calculate the vertical distance traveled by the ball.
We can use the kinematic equation for vertical displacement:
δy = V0yt + (1/2)gt^2

Plugging in the values:
δy = 15 × 1.53 + (1/2)(-9.8)(1.53)^2
δy ≈ 11.56 − 11.26
δy ≈ 0.3 m

Step 3: Calculate the height of the building.
The height of the building is equal to the vertical distance traveled by the ball before hitting the ground. Hence, the height will be:
Height = δy = 0.3 m

Therefore, the height of the building is approximately 0.3 meters.