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.

Answer 1

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

Answer 2

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.

Answer 3

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.