A projectile is launched horizontally from the edge of a cliff 20 meters high with an initial speed of 10 m/s.what is the horizontal distance the projectile travels before striking the level ground below the cliff?
To solve this problem, we need to use the kinematic equations of motion and consider the fact that the projectile is launched horizontally.
The vertical motion of the projectile can be analyzed using the equation:
y = y0 + v0yt - (1/2)gt²
Where:
- y is the vertical displacement (which is 0 in this case, since the projectile falls from the cliff to the ground);
- y0 is the initial vertical position (20 meters in this case);
- v0y is the initial vertical velocity (0 m/s in this case, since the projectile is launched horizontally);
- g is the acceleration due to gravity (approximately 9.8 m/s²);
- t is the time.
Since the projectile falls vertically, the time it takes to hit the ground can be found by solving the equation for t:
0 = 20 + 0t - (1/2)(9.8)t²
0 = -4.9t² + 20
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
In this case, a = -4.9, b = 0, and c = 20.
t = (-0 ± √(0² - 4(-4.9)(20))) / (2(-4.9))
t = ± √(0 + 392) / -9.8
t = ± √392 / -9.8
t ≈ ± 6.3 s
Since time cannot be negative in this case, we take t = 6.3 s.
Now, we can calculate the horizontal distance traveled by the projectile using the equation:
x = x0 + v0xt
Where:
- x is the horizontal displacement;
- x0 is the initial horizontal position (0 m in this case, since the projectile is launched from the edge of the cliff);
- v0x is the initial horizontal velocity (10 m/s in this case, since the projectile is launched horizontally);
- t is the time (6.3 s in this case).
x = 0 + 10 * 6.3
x ≈ 63 meters
Therefore, the horizontal distance the projectile travels before striking the level ground below the cliff is approximately 63 meters.