A projectile is launched at a 60 degree angle above the horizontal with an initial velocity of 35.5 m/s. What is the projectile's velocity (magnitude and direction) 1.50 s into the flight?

Vo = 35.5m/s @ 60o.

Xo = 35.5*cos60 = 17.75 m/s.
Yo = 35.5*sin60 = 30.74 m/s.

Y = Yo + g*t
Y = 30.74 - 9.8*1.5 = 16.04 m/s.

tanA = Y/Xo = 16.04/17.75 = 0.90366
A = 42.1o

V=Xo/cosA=17.75/cos42.1=23.9 m/[email protected].

To find the velocity (magnitude and direction) of the projectile 1.50 s into the flight, we need to split the initial velocity into its horizontal and vertical components.

The horizontal component of the initial velocity remains constant throughout the flight, while the vertical component is affected by gravity.

Given:
Initial velocity (v₀) = 35.5 m/s
Launch angle (θ) = 60 degrees
Time (t) = 1.50 s

Step 1: Splitting the Initial Velocity

The horizontal component (v₀x) can be found using the equation:
v₀x = v₀ * cos(θ)

The vertical component (v₀y) can be found using the equation:
v₀y = v₀ * sin(θ)

Substituting the given values, we have:
v₀x = 35.5 m/s * cos(60°)
v₀y = 35.5 m/s * sin(60°)

Horizontal component (v₀x) = 35.5 m/s * 0.5
Vertical component (v₀y) = 35.5 m/s * √3/2

Step 2: Finding the Velocity Components at t = 1.50 s

For the horizontal component (vₓ):
vₓ = v₀x

For the vertical component (vᵧ):
vᵧ = v₀y - g * t,

where g is the acceleration due to gravity (9.8 m/s²).

Substituting the given values, we have:
vᵧ = 35.5 m/s * √3/2 - (9.8 m/s²) * 1.50 s

Step 3: Finding the Velocity (Magnitude and Direction)

The magnitude of the velocity at a given time is given by the Pythagorean theorem:
v = √(vₓ² + vᵧ²)

The direction can be found using the inverse tangent function:
θ = tan^(-1)(vᵧ / vₓ)

Substituting the calculated values, we have:
v = √(vₓ² + vᵧ²)
θ = tan^(-1)(vᵧ / vₓ)

Calculate the values of v and θ to find the projectile's velocity (magnitude and direction) 1.50 s into the flight.