A 0.6 kg rock is projected from the edge of
the top of a building with an initial velocity of
11.1 m/s at an angle 51◦ above the horizontal.
Due to gravity, the rock strikes the ground at
a horizontal distance of 18.7 m from the base
of the building.
So, what is your thinking on this?
To find the time it takes for the rock to reach the ground, we first need to break down the initial velocity into its horizontal and vertical components. The horizontal component of velocity remains constant throughout the motion, while the vertical component is affected by gravity.
Given:
Mass of the rock (m) = 0.6 kg
Initial velocity (v) = 11.1 m/s
Angle of projection (θ) = 51°
Horizontal distance (d) = 18.7 m
Step 1: Find the horizontal component of velocity (v_x)
The horizontal component of velocity (v_x) is given by v_x = v * cos(θ).
v_x = 11.1 m/s * cos(51°)
v_x ≈ 11.1 m/s * 0.6293
v_x ≈ 6.9699 m/s
Step 2: Find the vertical component of velocity (v_y)
The vertical component of velocity (v_y) is given by v_y = v * sin(θ).
v_y = 11.1 m/s * sin(51°)
v_y ≈ 11.1 m/s * 0.7771
v_y ≈ 8.6076 m/s
Step 3: Find the time of flight (t)
The time of flight (t) is the duration it takes for the rock to reach the ground. We can calculate it using the vertical component of velocity and the acceleration due to gravity.
The equation of motion for vertical motion is:
d = v_y * t + 0.5 * g * t^2
Where:
d = the vertical displacement (height above the ground) = 0
v_y = vertical component of velocity = 8.6076 m/s
g = acceleration due to gravity ≈ 9.8 m/s^2
Since the rock lands on the ground, the vertical displacement (d) is 0. Solving the equation for time (t):
0 = 8.6076 t + 0.5 * 9.8 * t^2
4.9 t^2 + 8.6076 t = 0
Using the quadratic formula, we can solve for t:
t = (-8.6076 ± √(8.6076^2 - 4 * 4.9 * 0)) / (2 * 4.9)
t ≈ 0.9163 s (ignoring the negative solution)
So, the time it takes for the rock to reach the ground is approximately 0.9163 seconds.
Note: The result assumes there is no air resistance and a constant acceleration due to gravity.