Suppose you roll a bowling ball at 2.5 m/s across the roof of a flat building. It leaves the edge and strikes the ground 2.0 s later.
How high is the roof?
How far from the edge of the roof does the ball hit the ground?
Height = (1/2) g t^2
Distance from edge = 2.5 m/s * 2.0 s.
Well, let me put on my clown nose and calculate this for you!
To find the height of the roof, we can use the formula:
h = 1/2 * g * t^2
Where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time (2 seconds).
Plugging in the values, we get:
h = 1/2 * 9.8 m/s^2 * (2 seconds)^2
Calculating that, we find the height of the roof is approximately 19.6 meters.
Now, to find how far from the edge of the roof the ball hits the ground, we can use the formula for horizontal distance:
d = v * t
Where d is the distance, v is the initial velocity (2.5 m/s), and t is the time (2 seconds).
Plugging in the values, we get:
d = 2.5 m/s * 2 seconds
And that gives us a distance of approximately 5 meters.
So, the height of the roof is around 19.6 meters and the ball hits the ground about 5 meters from the edge of the roof. Just be glad you weren't standing under it!
To find the height of the roof, we can use the equation of motion:
h = (gt²)/2
Where:
h is the height of the roof,
g is the acceleration due to gravity (approximately 9.8 m/s²),
t is the time the ball takes to reach the ground.
Given that the time, t, is 2.0 seconds, we can substitute these values into the equation and solve for h:
h = (9.8 * 2.0²)/2
h = (9.8 * 4.0)/2
h = 19.6/2
h = 9.8 meters
Therefore, the height of the roof is 9.8 meters.
To find the distance from the edge of the roof where the ball hits the ground, we can use the equation of motion:
d = v * t
Where:
d is the distance,
v is the initial velocity of the ball, and
t is the time it takes for the ball to hit the ground.
Given that the velocity of the ball, v, is 2.5 m/s and the time, t, is 2.0 seconds, we can substitute these values into the equation and solve for d:
d = 2.5 * 2.0
d = 5.0 meters
Therefore, the ball will hit the ground 5.0 meters from the edge of the roof.
To find the height of the roof, we can use the formula for the vertical distance covered by an object under constant acceleration:
Δy = v₀t + 0.5at²
First, let's find the initial vertical velocity (v₀) of the ball as it leaves the edge. We know that the ball is rolling horizontally, so its initial vertical velocity is 0 m/s.
Next, we need to find the time it takes for the ball to reach the ground. We are told that this time is 2.0 seconds (t).
Since the ball falls freely under gravity, the acceleration (a) is equal to the acceleration due to gravity, which is approximately 9.8 m/s².
Substituting the given values into the formula, we have:
Δy = (0)(2.0) + 0.5(9.8)(2.0)
= 0 + (9.8)(1.0)
= 9.8 meters
Therefore, the height of the roof is 9.8 meters.
To find the horizontal distance from the edge of the roof to where the ball hits the ground, we can use the formula for horizontal distance:
Δx = v₀x * t
Since the ball was rolling horizontally with a velocity of 2.5 m/s, the initial horizontal velocity (v₀x) is equal to 2.5 m/s.
Substituting the given values into the formula, we have:
Δx = (2.5)(2.0)
= 5.0 meters
Therefore, the ball hits the ground 5.0 meters from the edge of the roof.