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?
The ball lands 140m from the base of the building. The balls initial horizontal velocity is 20m/s.
140m = (20m/s) * t
Solve for t.
Since the time the ball is in flight is now known.
Plug that into the "complete" formula for distance:
d = (1/2)*(g)*(t^2) + v*t + h
Let distance 0 be ground level.
initial vertical velocity is v=15m/s
h is the initial distance above the ground which is the height of the building.
0 = (1/2)*(g)*(t^2) + v*t + h
Solve for h.
Remember that the initial velocity is up and the acceleration of gravity is down (watch the signs)
I helped you with this question yesterday and gave the exact same info as Quidditch Why didn't you use the info I gave?
Thanks! I saw it and used it but I'm not sure if I did it right. I got 140 for the height of the building and I don't think that's right.
I got something in that range, but I believe your answer has more error than I would expect.
What did you get for time that the ball was in the air?
What numbers did you plug into your final equation?
For time t = d/v
140/20
t= 7
0 = 1/2(-10)(7^2)+(15)(7)+h
-h = -140
h = 140
We were asked to use 10 for our acceleration.
You got it!
Using 10m/(s^2) for gravity, your answer is correct.
To calculate the height of the building, you'll need to use the equations of motion. The first step is to separate the horizontal and vertical components of the motion.
Let's start by calculating the time it takes for the ball to hit the ground. The vertical motion follows a free-fall trajectory, so we can use the equation:
y = V1y * t + (1/2) * a * t^2
where:
y = vertical displacement (height of the building)
V1y = initial vertical velocity (15 m/s)
t = time
a = acceleration due to gravity (-9.8 m/s^2)
Since the ball hits the ground, the vertical displacement will be equal to zero. Substituting these values into the equation, we get:
0 = 15 * t + (1/2) * (-9.8) * t^2
This equation is a quadratic equation in terms of 't'. Solve it to find the time it takes for the ball to hit the ground. You can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
where:
a = (-9.8/2)
b = 15
c = 0
Once you find the value of 't', you can substitute it into the horizontal equation to find the horizontal displacement:
x = V1x * t
where V1x is the initial horizontal velocity (20 m/s).
From the problem statement, we know that the horizontal displacement is 140 m. So we can solve for 't' using the equation:
140 = 20 * t
Once you've found the value of 't', substitute it back into the vertical equation to find the height of the building:
y = 15 * t + (1/2) * (-9.8) * t^2
This will give you the height of the building.