A ball is thrown at an angle of 37.1degrees above the horizontal from a point on the top of a building and 60m from the edge of the building. It just misses the edge of the building and continues on towards the ground. The building is 50m high. At what speed was the ball thrown and how far from the vertical wall of the building does the ball land?

Write an equation that says that the horozontal distance travelled during the time the ball is above the top of the building equals 60 m. The only unknown will be the launch velocity V.

The time T spent in the air above the top of the building is given by:
V sinA = g T/2
T = 2 V sinA/g
The horizontal distance travelled above the building is
VcosA*T = 2 V^2 sin A cos A /g = 60 m
V^2 sin (2A)/g = 60 m
where A = 37.1 degrees.

Solve for V

V cos A * T' is the distance the ball lands away from the building; T' is the additional time it takes to get to the bottom, 50 m below. Remember that it will start out with a downward velocity component -V sin A when it passes the edge of the building

A ball is thrown at an angle of 37.1 degrees above the horizontal from a point on the top of a building and 60m from the edge of the building. It just misses the edge of the building and continues on towards the ground. The building is 50m high. At what speed was the ball thrown and how far from the vertical wall of the building does the ball land?

The horizontal distance a projectile travels is derivable from d = Vo^2sin(2µ)/g where d is the horizontal distance in meters(same elevation), Vo is the initial velocity in m/s, g is the acceleration due to gravity = 9.8m/sec.^2 and µ is the elevation of the initial velocity direction in degrees.

Solving for Vo, Vo = sqrt(9.8(60)/sin(74.2º)) = 24.72m/s.

The ball will clear the edge of the building with same speed as it left the launch point and at the same angle , but below the horizontal.

The horizontal component, Vh, of the velocity at that point is 24.72(cos(37.1) = 19.716m/s and the vertical component, Vv, is 24.72(sin(37.1)) = 14.911m/s.

The time to hit the ground from the edge of the building is derivable from h = Vvt + 9.8t^2 where h = 50m, or 50 = 14.911t + 9.8t^2 or 9.8t^2 + 14.911t - 50 = 0. From the quadratic formula, t = 1.622 sec.

The impact distance from the building is therefore d = tVh = 1.622(19.716) = 31.98m.

To solve this problem, we can break it down into two separate parts: analyzing the horizontal motion and the vertical motion of the ball.

Step 1: Analyzing horizontal motion:
We know that the initial horizontal displacement of the ball is 60m, and it doesn't change during the entire motion. Therefore, the horizontal velocity of the ball remains constant throughout.

Step 2: Analyzing vertical motion:
We can break down the motion into its vertical and horizontal components. The vertical motion can be analyzed using the equations of uniformly accelerated motion.

Given data:
- Initial vertical displacement (height of the building) = 50m
- Angle of projection = 37.1 degrees

Step 2.1: Resolve the initial velocity into vertical and horizontal components:
The initial velocity (V₀) of the ball can be resolved into vertical (V₀y) and horizontal (V₀x) components using trigonometry.

V₀y = V₀ * sin θ
V₀x = V₀ * cos θ

Step 2.2: Analyze vertical motion:
We can use the equations of motion to analyze the vertical motion of the ball. The equation we need is:
h = V₀y * t - 0.5 * g * t²

In this case, the vertical displacement (h) is -50m (negative value since we are measuring the height from the top of the building downwards). The acceleration due to gravity (g) is approximately -9.8 m/s² (negative value because it acts downward).

-50 = V₀ * sin θ * t - 0.5 * (-9.8) * t²
-50 = V₀ * sin θ * t + 4.9t²

Step 2.3: Calculate the time of flight (t):
To find the time it takes for the ball to reach the ground, we need to determine when the vertical displacement becomes zero (since it reaches the ground).

0 = V₀ * sin θ * t + 4.9t²
V₀ * sin θ * t = -4.9t²
V₀ * sin θ = -4.9t
t = -(V₀ * sin θ) / 4.9

Step 2.4: Calculate the initial speed (V₀):
To find the initial speed, we can use the horizontal displacement (60m) and the time of flight (t) determined in step 2.3.

V₀x = V₀ * cos θ
V₀ = 60 / (t * cos θ)

Step 3: Calculate the distance from the vertical wall:
To determine the distance from the vertical wall, we need to find the horizontal distance covered by the ball during the time of flight (t).

Distance from the wall = V₀x * t

Summary of the steps to find the speed and distance:
1. Resolve the initial velocity into vertical and horizontal components.
2. Analyze the vertical motion to find the time of flight (t).
3. Find the initial speed (V₀) using the horizontal displacement and time of flight.
4. Calculate the distance from the vertical wall using the horizontal velocity and time of flight.

Note: To calculate the exact numerical values, we'll need to know the value of the initial speed (V₀).

To solve this problem, we can break it down into two components: horizontal motion and vertical motion. Let's start with the horizontal motion of the ball.

First, we need to find the time it takes for the ball to travel 60m horizontally. We can use the horizontal component of the velocity of the ball to find this time.

Horizontal velocity (Vx) = initial speed (V0) * cos(angle)
Vx = V0 * cos(37.1°)

We also know that the horizontal displacement (x) is 60m. Using the equation x = Vx * t, where t is the time taken, we can rearrange the equation to solve for t:

t = x / Vx
t = 60m / (V0 * cos(37.1°))

Next, let's move on to the vertical motion of the ball.

Considering the vertical motion, we can determine the initial vertical velocity (Vy) using the initial speed (V0) and the angle of projection.

Vertical velocity (Vy) = initial speed (V0) * sin(angle)
Vy = V0 * sin(37.1°)

We are given the height of the building (50m), so we can use the kinematic equation h = Vy * t - (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.

Simplifying this equation, we get:
50m = (V0 * sin(37.1°)) * t - (1/2) * 9.8 m/s^2 * t^2

Now, we have two equations with two unknowns (V0 and t). We can solve these equations simultaneously to find the values.

t = 60m / (V0 * cos(37.1°)) --------------- Equation 1
50m = (V0 * sin(37.1°)) * t - (1/2) * 9.8 m/s^2 * t^2 --------------- Equation 2

By substituting Equation 1 into Equation 2, we can solve for V0. Once we find V0, we can substitute it back into Equation 1 to solve for t.

Once we have the values of V0 and t, we can find the distance from the vertical wall of the building where the ball lands by multiplying the horizontal component of the velocity (Vx) by the time (t):

Distance from the wall = Vx * t.

Now, let's solve these equations to find the answers.