A stone is thrown upward from the edge of a cliff overlooking the ocean. The initial speed of the stone is 20m/s, and it's thrown up at an angel of 30 degrees with respect to the horizontal. the cliff is 45m above the ocean surface. How long is the stone in flight before striking the water?
in the vertical:
hf=hi+vi*t-4.9t^2
0=45+20*sin30*t-4.9t^2
put that in standard form for a quadratic
at^2+bt+c=0
use the quadratic equation
t=(-b+-sqrt(b^2-4ac))/2a
To find the time the stone is in flight before striking the water, we can use the equations of motion.
Step 1: Break the initial velocity of the stone into its horizontal and vertical components.
Given:
Initial speed (u) = 20 m/s
Launch angle (θ) = 30 degrees
The horizontal component of the initial velocity (u_x) is given by:
u_x = u * cos(θ)
The vertical component of the initial velocity (u_y) is given by:
u_y = u * sin(θ)
Substituting the given values:
u_x = 20 * cos(30) ≈ 17.32 m/s
u_y = 20 * sin(30) ≈ 10 m/s
Step 2: Find the time taken for the stone to reach the highest point of its trajectory.
In this case, the vertical component of the velocity becomes zero at the highest point. We can use the following equation:
v_y = u_y + a_y * t
Since the stone is moving upward, the acceleration in the vertical direction (a_y) is equal to the acceleration due to gravity (-9.8 m/s²).
At the highest point, v_y = 0, so we can rewrite the equation as:
0 = 10 - 9.8 * t
Solving for t:
t = 10 / 9.8 ≈ 1.02 seconds
Step 3: Determine the total time of flight.
Since the stone reaches the highest point in half of the total time of flight, we can multiply the time obtained in step 2 by 2:
Total time of flight = 2 * 1.02 ≈ 2.04 seconds
Therefore, the stone is in flight for approximately 2.04 seconds before striking the water.