Question

The range is the horizontal distance from the cannon when the pumpkin hits the ground. This distance is given by the product of the horizontal velocity (which is constant) and the amount of time the pumpkin is in the air (which is determined by the vertical component of the initial velocity, as you just discovered). Set the initial speed to 14 m/s, and fire the pumpkin several times while varying the angle between the cannon and the horizontal. For which angle is the range a maximum (with the initial speed held constant)

Answer 1

x(max) = 20 m

Step-by-step explanation:

In projectile shoot e have:

V₀y = V₀*sin θ

V₀ₓ = V₀*cos θ

V₀ initial speed

θ shotting angle

Vₓ = V₀ₓ = V₀*cos θ during the trajectory

Vy = V₀y - g*t

And when Vy = 0 h is maximun and the time to reach y maximum is half of the overall time.

According to that

Vy = 0 V₀y = g*t t = V₀y/g

t = 14* sinθ / 9,8

t = 1,43 *sinθ s

And overall time is T = 2* 1,43*sinθ

t = 2,86*sinθ

x = V₀ₓ * t s

x = 14*cosθ * 2,86* sinθ

x = 40 * cosθ * sinθ

Lets take two very well know sin cos pairs

for ∡ 30 ∡ 45 ∡60

and 30 and 60 are complementary angles for boths

cosθ * sinθ is the same ( 1/2)*(√3/2)

cosθ * sinθ = 0,433 and

x = 40* 0,433

x = 17,32 m

For ∡ 45 boths sin45 and cos45 are equal √2/2

In this case

x = 40*(√2/2)*(√2/2)

x = 40* 2/4

x = 20 m

And that is the maximum range

x(max) = 20 m