a ball is launched from the top of a 23 m high cliff with a velocity of 33 m/s oriented 12 degrees above the horizonal. Determine the time it takes for the ball to hit the ground.
the vertical launch velocity is
... 33 sin(12º)
0 = 1/2 g t² + [33 sin(12º)] t + 23
use the quadratic formula to find t
...you want the positive solution
g = 9.8 m/s²
To determine the time it takes for the ball to hit the ground, we can use the concept of projectile motion. We will break down the initial velocity of the ball into its vertical and horizontal components.
Given:
Initial height of the cliff, h = 23 m
Initial velocity of the ball, v₀ = 33 m/s
Launch angle, θ = 12 degrees
First, let's find the vertical component of the initial velocity (v₀_y):
v₀_y = v₀ * sin(θ)
v₀_y = 33 * sin(12°)
v₀_y ≈ 6.95 m/s
Next, we can find the time it takes for the ball to reach the maximum height (t_max) using the equation:
v_y = v₀_y - g * t_max, where g is the acceleration due to gravity (approximately 9.8 m/s²).
At the highest point, the vertical velocity becomes zero, so:
0 = v₀_y - g * t_max
t_max = v₀_y / g
t_max = 6.95 / 9.8
t_max ≈ 0.711 seconds
Now, to find the total time of flight, we can use the equation:
h = v₀_y * t_total - 0.5 * g * t_total²
Since the ball lands on the ground, the final height (h) is zero:
0 = v₀_y * t_total - 0.5 * g * t_total²
0 = 6.95 * t_total - 0.5 * 9.8 * t_total²
Simplifying the equation:
4.9 * t_total² - 6.95 * t_total = 0
This equation can be solved by factoring or using the quadratic formula. By factoring, we get:
t_total (4.9 * t_total - 6.95) = 0
This gives two possible solutions:
t_total = 0 (not possible since the ball has been launched)
t_total = 6.95 / 4.9 ≈ 1.418 seconds
Since the ball is in the air for the same amount of time on its way up and down, we can divide the total time by 2 to find the time it takes for the ball to hit the ground:
t_ground = t_total / 2
t_ground ≈ 1.418 / 2 ≈ 0.709 seconds
Therefore, the time it takes for the ball to hit the ground is approximately 0.709 seconds.