Answer:
The speed of the rock when it strikes the ground is 34.55 m/s
Explanation:
Given;
height of the building, h = 15 m
initial velocity of the rock, V₀ = 30 m/s
angle of projection, θ = 33°
The velocity of the rock before it strikes the ground, can be calculated from vertical component of the velocity and horizontal component of the velocity.
Vertical component of the velocity,[tex]V_y[/tex]
[tex]V_y^2 = (V_oSin \theta)^2 + 2gh\\\\V_y^2 = (30Sin \ 33)^2 + 2(9.8)(15)\\\\V_y^2 = 266.93 + 294\\\\V_y = \sqrt{560.93} \\\\V_y = 23.684 \ m/s[/tex]
Horizontal component of the velocity, [tex]V_x[/tex]
[tex]V_x = V_0 Cos\theta\\\\V_x = 30Cos33\\\\V_x = 25.161 \ m/s[/tex]
The velocity of the rock when it strikes the ground, V
[tex]V = \sqrt{V_y^2 + V_x^2} \\\\V = \sqrt{23.684^2 + 25.161^2} \\\\V = 34.55 \ m/s[/tex]
Therefore, the speed of the rock when it strikes the ground is 34.55 m/s
