Question

A stone is tossed horizontally from the highest point of a 95 m building and lands 105 m from the base of the building. Ignore air resistance, and use a coordinate system whose origin is at the highest point of the building, with positive y upwards and positive x in the direction of the throw.
A. How long is the stone in the air in s?
B. What must have been the initial horizontal component of the velocity, in m/s?
C. What is the vertical component of the velocity just before the stone hits the ground, in m/s?
D. What is the magnitude of the velocity of the stone just before it hits the ground, in m/s?

Answer 1

Answer:

A) t = 4.40 s , B)   v = 23.86 m / s ,  c)  v_y = - 43.12 m / s , D)  v = 49.28 m/s

Explanation:

This is a projectile throwing exercise,

A) To know the time of the stone in the air, let's find the time it takes to reach the floor

          y = y₀ + $v_{oy}$ t - ½ g t²

as the stone is thrown horizontally  v_{oy} = 0

          y = y₀ - ½ g t²

          0 = y₀ - ½ g t²

          t = √ (2 y₀ / g)

          t = √ (2 95 / 9.8)

          t = 4.40 s

B) what is the horizontal velocity of the body

          v = x / t

          v = 105 / 4.40

          v = 23.86 m / s

C) The vertical speed when it touches the ground

          v_y = $v_{oy}$ - g t

          v_y = 0 - 9.8 4.40

          v_y = - 43.12 m / s

the negative sign indicates that the speed is down

D) total velocity just hitting the ground

          v = vₓ i ^ + v_y j ^

          v = 23.86 i ^ - 43.12 j ^

Let's use Pythagoras' theorem to find the modulus

          v = √ (vₓ² + v_y²)

          v = √ (23.86² + 43.12²)

           v = 49.28 m / s

we use trigonometry for the angle

          tan θ = v_y / vₓ

          θ = tan⁻¹ (-43.12 / 23.86)

            θ = -61