A boy throws a ball at 5.5 m/s at an angle of 25.4° above the horizontal off a cliff which is 70.8 m high. What angle does the velocity of the ball make with the horizontal 0.78 s after the throw? (make the angle positive if the angle is above the horizontal, and negative if it is below the horizontal)

To find the angle that the velocity of the ball makes with the horizontal 0.78 s after the throw, we need to break down the given information and solve step by step.

Let's start by finding the horizontal and vertical components of the initial velocity of the ball.

Given:
Initial velocity (magnitude): 5.5 m/s
Launch angle: 25.4°

Horizontal component of velocity (Vx):
Vx = V_initial * cos(angle)
= 5.5 m/s * cos(25.4°)

Vertical component of velocity (Vy):
Vy = V_initial * sin(angle)
= 5.5 m/s * sin(25.4°)

Now, we can use the vertical motion equation to determine the time it takes for the ball to reach the top of the cliff.

Given:
Height of cliff: 70.8 m
Vertical acceleration due to gravity: -9.8 m/s² (negative because it acts in the opposite direction)

Using the equation:
Δy = V_initialy * t + (1/2) * a * t²

Where:
Δy = change in vertical position (70.8 m)
V_initialy = initial vertical velocity (Vy)
t = time
a = acceleration due to gravity (-9.8 m/s²)

Solving for t:
70.8 m = (5.5 m/s * sin(25.4°)) * t + (1/2) * (-9.8 m/s²) * t²

This equation is a quadratic equation, so we can solve it using the quadratic formula or by factoring. For simplicity, let's use the quadratic formula:

t² + (2 * (5.5 m/s * sin(25.4°)) / (-9.8 m/s²)) * t - (2 * 70.8 m / (-9.8 m/s²)) = 0

Solve for t using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)

In this case:
a = 1
b = (2 * (5.5 m/s * sin(25.4°)) / -9.8 m/s²)
c = (2 * 70.8 m / -9.8 m/s²)

Once you have the value of t, which should be around 0.78 seconds, we can move to the next step.

To find the angle that the velocity of the ball makes with the horizontal at 0.78 seconds, we need to find the new horizontal and vertical components of the velocity.

Horizontal component of velocity at 0.78 s (Vx'):
The horizontal component of velocity remains constant throughout the motion, so Vx' is the same as Vx calculated earlier.

Vertical component of velocity at 0.78 s (Vy'):
Using the equation:
Vy' = Vy + a * t

Where:
a = acceleration due to gravity (-9.8 m/s²)
Vy = initial vertical velocity (Vy)
t = time (0.78 s)

Vy' = (5.5 m/s * sin(25.4°)) + (-9.8 m/s²) * 0.78 s

Now, we have Vx' and Vy' at 0.78 seconds. To find the angle, θ', with the horizontal, we can use the tangent function:

θ' = arctan(Vy' / Vx')

Calculate the value of Vy' / Vx', then take the arctan to find θ'. The resulting angle will be positive if it is above the horizontal and negative if it is below the horizontal.