A spring gun launches a ball horizontally from 1.3 m above the floor. the ball strikes the floor 2.2m away, measured horizontally. Calculate

A .. the time the ball is flight
B... the vertical component of its velocity at the time it strikes the floor
C.... the initial velocity Vo ( magnitude and direction).
D...the horizontal component of its velocity as it hits the floor

The initial horizontal velocity component is Vo. It equals the horizontal component when it his the floor.

(a) The time to fall H = 1.3 m, with no initial velocity component, is
T = sqrt(2H/g) = 0.515 s

(b) Vy at t = T = -g*T

c) The initial direction is horizontal. They tell you that. Vo = Vx = (2.2 m)/T = 4.27 m/s

d) Vx remains Vo

To calculate the various quantities related to the motion of the ball, we can use the equations of motion. Let's break down each part of the problem step by step:

A. Calculate the time the ball is in flight:
Since the ball is launched horizontally, there is no vertical acceleration acting on it. Therefore, we only need to consider the vertical motion of the ball to calculate the time of flight.
Using the equation for vertical displacement:
Δy = V₀y * t + (1/2) * g * t²

In this case, Δy (vertical displacement) is -1.3 m (negative because the ball is falling), V₀y (initial vertical velocity) is equal to zero (since the ball is launched horizontally), and g (acceleration due to gravity) is -9.8 m/s² (negative because it acts downward).
Simplifying the equation, we get:
-1.3 = (1/2) * -9.8 * t²

Solving for t, we find:
t² = (2 * 1.3) / 9.8
t ≈ sqrt(0.2653)
t ≈ 0.5151 seconds

Therefore, the time the ball is in flight is approximately 0.5151 seconds.

B. Calculate the vertical component of its velocity at the time it strikes the floor:
The vertical component of velocity at any time can be found using the equation:
Vy = V₀y + g * t

Since the ball is launched horizontally, the initial vertical velocity (V₀y) is zero. Substituting the values, we get:
Vy = 0 + (-9.8) * 0.5151

Therefore, the vertical component of velocity at the time the ball strikes the floor is approximately -5.0429 m/s (negative because it is directed downward).

C. Calculate the initial velocity (V₀) magnitude and direction:
To calculate the magnitude of the initial velocity, we can use the horizontal displacement and time of flight.
Using the equation for horizontal displacement:
Δx = V₀x * t

In this case, Δx (horizontal displacement) is 2.2 m, V₀x (initial horizontal velocity) is unknown, and t (time of flight) is 0.5151 seconds.

Rearranging the equation, we get:
V₀x = Δx / t
V₀x = 2.2 / 0.5151
V₀x ≈ 4.2736 m/s

To find the magnitude of the initial velocity vector (V₀), we can use the Pythagorean theorem:
V₀ = sqrt(V₀x² + V₀y²)
V₀ = sqrt((4.2736)² + 0²)

Therefore, the magnitude of the initial velocity is approximately 4.2736 m/s.

The direction of the initial velocity can be obtained by finding the inverse tangent of V₀y/V₀x:
θ = atan(V₀y / V₀x)
θ = atan(0 / 4.2736)

Since V₀y is zero, the direction of the initial velocity is horizontal.

D. Calculate the horizontal component of its velocity as it hits the floor:
Since the ball's initial horizontal velocity remains constant throughout its flight, the horizontal component of velocity at the time it hits the floor is the same as the initial horizontal velocity.
Therefore, the horizontal component of velocity as the ball hits the floor is approximately 4.2736 m/s.