height displacement

Question

1.3 A stone is thrown at a velocity of42 m/s at an angle of26°to the horizontal. Calculate the following: 1.3.1 The maximum height that the stone reaches 1.3.2 The horizontal displacement of the stone

Answer 1

1.3.1)

Find the maximum height of the projectile:

H = v₀² sin²θ / 2g

Here, θ is the angle with horizontal, v₀ is the initial projectile velocity.

Substitute 42 m/s for v_0, 26° for θ, and 9.81 m/s² for g.

H = (42)² sin²26° / 2(9.81)

  = 17.278 m

Answer 2

1.3.2)

Find the horizontal displacement of the stone of the projectile:

R = v₀² sin2θ / g

Here, θ is the angle with horizontal, v₀ is the initial projectile velocity.

Substitute 42 m/s for v_0, 26° for θ, and 9.81 m/s² for g.

R = (42)² sin2(26°) / (9.81)

  = 141.697 m