Question
A ball is thrown at 27° below the horizontal from a rooftop of height 90 m. It lands 2.1 s later. Find the point of impact with the ground from the base of the building if the initial speed is 71.73529599m/s.
Answers
Answered by
oobleck
the height of the ball at time t is
h = 90 - 71.73529599 sin27° t - 4.9t^2
so h=0 at t=2.1
x = 2.1 * 71.73529599 cos27° = 134.225
h = 90 - 71.73529599 sin27° t - 4.9t^2
so h=0 at t=2.1
x = 2.1 * 71.73529599 cos27° = 134.225
Answered by
GPT 3.5
To solve this problem, we can split the initial velocity into its horizontal and vertical components.
The horizontal component of velocity can be found using the formula:
Vx = V * cos(theta)
where Vx is the horizontal component of velocity, V is the initial speed, and theta is the angle of projectile motion (27° below the horizontal).
Vx = 71.73529599 m/s * cos(27°)
Vx = 65.3092 m/s
The vertical component of velocity can be found using the formula:
Vy = V * sin(theta)
where Vy is the vertical component of velocity, V is the initial speed, and theta is the angle of projectile motion (27° below the horizontal).
Vy = 71.73529599 m/s * sin(27°)
Vy = 34.4607 m/s
The time of flight can be found using the formula:
t = (2 * Vy) / g
where t is the time of flight, Vy is the vertical component of velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
t = (2 * 34.4607 m/s) / 9.8 m/s^2
t = 7.066 s
The horizontal distance traveled by the ball can be found using the formula:
dx = Vx * t
where dx is the horizontal distance, Vx is the horizontal component of velocity, and t is the time of flight.
dx = 65.3092 m/s * 2.1 s
dx = 137.149 m
The point of impact with the ground from the base of the building is therefore approximately 137.149 meters.
The horizontal component of velocity can be found using the formula:
Vx = V * cos(theta)
where Vx is the horizontal component of velocity, V is the initial speed, and theta is the angle of projectile motion (27° below the horizontal).
Vx = 71.73529599 m/s * cos(27°)
Vx = 65.3092 m/s
The vertical component of velocity can be found using the formula:
Vy = V * sin(theta)
where Vy is the vertical component of velocity, V is the initial speed, and theta is the angle of projectile motion (27° below the horizontal).
Vy = 71.73529599 m/s * sin(27°)
Vy = 34.4607 m/s
The time of flight can be found using the formula:
t = (2 * Vy) / g
where t is the time of flight, Vy is the vertical component of velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
t = (2 * 34.4607 m/s) / 9.8 m/s^2
t = 7.066 s
The horizontal distance traveled by the ball can be found using the formula:
dx = Vx * t
where dx is the horizontal distance, Vx is the horizontal component of velocity, and t is the time of flight.
dx = 65.3092 m/s * 2.1 s
dx = 137.149 m
The point of impact with the ground from the base of the building is therefore approximately 137.149 meters.
Answered by
GPT 3.5
I apologize for the incorrect calculation in my previous response.
Using the correct formula for the height at time t and considering that h = 0 at t = 2.1 s, we can calculate the correct horizontal distance.
The formula for the height at time t is:
h = 90 - 71.73529599 * sin(27°) * t - 4.9 * t^2
Setting h = 0 and solving for t, we get:
0 = 90 - 71.73529599 * sin(27°) * t - 4.9 * t^2
0 = 4.9 * t^2 - 71.73529599 * sin(27°) * t - 90
Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 4.9, b = -71.73529599 * sin(27°), and c = -90.
Plugging these values into the formula, we get:
t = (-(-71.73529599 * sin(27°)) ± √((-71.73529599 * sin(27°))^2 - 4 * 4.9 * (-90))) / (2 * 4.9)
Simplifying and calculating, we find two possible values for t:
t ≈ 2.045 s or t ≈ 10.27 s
Since we are interested in the time t = 2.1 s when the ball lands, we can conclude that the ball is at its highest point above the ground at t ≈ 10.27 s. At t = 2.1 s, the ball has landed.
To find the horizontal distance traveled by the ball, we can use the formula:
x = 2.1 * 71.73529599 * cos(27°)
x ≈ 134.225 meters
So, from the base of the building, the point of impact with the ground is approximately 134.225 meters away.
Using the correct formula for the height at time t and considering that h = 0 at t = 2.1 s, we can calculate the correct horizontal distance.
The formula for the height at time t is:
h = 90 - 71.73529599 * sin(27°) * t - 4.9 * t^2
Setting h = 0 and solving for t, we get:
0 = 90 - 71.73529599 * sin(27°) * t - 4.9 * t^2
0 = 4.9 * t^2 - 71.73529599 * sin(27°) * t - 90
Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 4.9, b = -71.73529599 * sin(27°), and c = -90.
Plugging these values into the formula, we get:
t = (-(-71.73529599 * sin(27°)) ± √((-71.73529599 * sin(27°))^2 - 4 * 4.9 * (-90))) / (2 * 4.9)
Simplifying and calculating, we find two possible values for t:
t ≈ 2.045 s or t ≈ 10.27 s
Since we are interested in the time t = 2.1 s when the ball lands, we can conclude that the ball is at its highest point above the ground at t ≈ 10.27 s. At t = 2.1 s, the ball has landed.
To find the horizontal distance traveled by the ball, we can use the formula:
x = 2.1 * 71.73529599 * cos(27°)
x ≈ 134.225 meters
So, from the base of the building, the point of impact with the ground is approximately 134.225 meters away.
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