A projectile is fired from the ground level with initial velocity of 90m/s at an angle of 36degre from horizontal. find the magnitude and direct of projectile 5 secs after firring

To find the magnitude and direction of a projectile 5 seconds after firing, we can treat the horizontal and vertical components of motion separately.

First, let's break down the initial velocity of the projectile into its horizontal and vertical components. We can use the following trigonometric equations:

Horizontal component: Vx = V₀ * cos(θ)
Vertical component: Vy = V₀ * sin(θ)

Where:
Vx is the horizontal component of velocity,
Vy is the vertical component of velocity,
V₀ is the initial velocity (90 m/s in this case),
and θ is the angle of projection (36 degrees in this case).

Plugging in the values, we get:

Vx = 90 m/s * cos(36°)
Vx ≈ 90 m/s * 0.809
Vx ≈ 72.81 m/s

Vy = 90 m/s * sin(36°)
Vy ≈ 90 m/s * 0.5878
Vy ≈ 52.91 m/s

Now, let's find the magnitude and direction of the projectile 5 seconds after firing. Since we know the initial velocity components, we can use the following equations of motion:

Horizontal displacement: Dx = Vx * t
Vertical displacement: Dy = Vy * t - (1/2) * g * t²

Where:
Dx is the horizontal displacement,
Dy is the vertical displacement,
t is the time (5 seconds in this case),
and g is the acceleration due to gravity (approximately 9.8 m/s²).

Plugging in the values and solving, we get:

Dx = 72.81 m/s * 5 s
Dx = 364.05 m

Dy = 52.91 m/s * 5 s - (1/2) * 9.8 m/s² * (5 s)²
Dy = 264.55 m

To find the magnitude of the projectile's position, we can use the Pythagorean theorem:

Magnitude = √(Dx² + Dy²)
Magnitude = √(364.05² + 264.55²)
Magnitude ≈ 450.69 m

The direction of the projectile can be found using the tangent function:

tan(θ) = Dy / Dx
θ = arctan(Dy / Dx)

θ = arctan(264.55 m / 364.05 m)
θ ≈ 35.05°

Therefore, the magnitude of the projectile 5 seconds after firing is approximately 450.69 m, and the direction is approximately 35.05 degrees above the horizontal.