A body is projected with a velocity 80m/s at 30 with the horizontal. What is its range and greatest height realized? (g=10m/s²)

Answer 1

To find the range and greatest height realized, we can break down the initial velocity into its vertical and horizontal components.

The vertical component of the initial velocity (Vy) can be calculated using the formula Vy = V * sin(θ), where V is the initial velocity and θ is the angle of projection. In this case, V = 80 m/s and θ = 30°.

Vy = 80 m/s * sin(30°)
Vy = 80 m/s * 0.5
Vy = 40 m/s

The horizontal component of the initial velocity (Vx) can be calculated using the formula Vx = V * cos(θ), where V is the initial velocity and θ is the angle of projection.

Vx = 80 m/s * cos(30°)
Vx = 80 m/s * √3/2
Vx = 40 m/s * √3

Now let's calculate the time of flight (t) using the formula t = 2 * Vy / g, where g is the acceleration due to gravity (10 m/s²).

t = 2 * 40 m/s / 10 m/s²
t = 8 s

The range (R) can be calculated using the formula R = Vx * t, where Vx is the horizontal component of the initial velocity and t is the time of flight.

R = 40 m/s * √3 * 8 s
R = 320 √3 m

The greatest height (H) can be calculated using the formula H = Vy^2 / (2 * g), where Vy is the vertical component of the initial velocity and g is the acceleration due to gravity.

H = (40 m/s)^2 / (2 * 10 m/s²)
H = 1600 m²/s² / 20 m²/s²
H = 80 m

Therefore, the range of the projectile is approximately 320 √3 meters and the greatest height realized is 80 meters.