Ball is thrown horizontally off a cliff with an initial speed of 20m/s. a) how long does it take the ball to hit the ground 20m below. b)how long would it take to hit the ground if it was dropped from the cliff? c) how long would it take to hit the ground if it was thrown with an intial velocity of 20m/s at 18 degrees below the horizontal?
Not for sure if im doing this right.
For a) I used: change in x=vi*t; t=20m/20m/s
a) 1sec?
b) change in y=vi(of y) *t-1/2(g)(t^2)
20m=0*t-1/2(9.8)(t^2)
20m=4.9t^2
t^2=4.08
t=2.02 seconds
c) got 1.05 seconds, but that doesnt seem right to me, considering my answer for a)
Any help would be appreciated, especially if you can tell me where I'm going wrong.
good waaaannnn
To solve these problems, you need to use the kinematic equations for projectile motion. In this case, since the horizontal velocity is constant and the vertical velocity is affected by gravity, you can separate the horizontal and vertical components of the motion.
a) The ball is thrown horizontally, so the initial vertical velocity is zero. The vertical displacement is given as 20m below the cliff. The acceleration due to gravity is -9.8m/s² (negative because it acts downward). You can use the equation for vertical displacement:
change in y = viy * t + (1/2) * a * t^2
Since viy = 0, the equation simplifies to:
20m = 0 * t + (1/2) * (-9.8m/s²) * t^2
Simplifying further:
20m = -4.9m/s² * t^2
Now, rearrange the equation to solve for t^2:
t^2 = 20m / (-4.9m/s²)
t^2 ≈ -4.08s²
Since time cannot be negative, there seems to be an error in your calculations for part a. Let's correct it:
t ≈ √(20m / (-4.9m/s²))
t ≈ 2.02 seconds (rounded to two decimal places)
So, the ball takes approximately 2.02 seconds to hit the ground 20m below when thrown horizontally off the cliff.
b) If the ball is dropped from the cliff, the initial vertical velocity is still zero, but this time, the initial height (change in y) is also zero. You can use the same equation for vertical displacement as in part a:
change in y = viy * t + (1/2) * a * t^2
0 = 0 * t + (1/2) * (-9.8m/s²) * t^2
Simplifying:
0 = -4.9m/s² * t^2
Since displacement is zero, the ball is already at the ground (no vertical displacement). This means the equation reduces to:
0 = -4.9m/s² * t^2
This equation implies that time is irrelevant; the ball hits the ground instantaneously when dropped. Therefore, the time it takes to hit the ground when dropped from the cliff is zero.
c) In this case, the ball is thrown with an initial velocity of 20m/s at an angle of 18 degrees below the horizontal. You need to consider both the horizontal and vertical components of the motion.
The horizontal velocity component (vix) can be found using the equation: vix = vi * cos(θ). Substituting the values:
vix = 20m/s * cos(18°)
The vertical velocity component (viy) can be found using the equation: viy = vi * sin(θ). Substituting the values:
viy = 20m/s * sin(18°)
Now you can follow the same steps as in part a to determine the time it takes for the ball to hit the ground with these components:
change in y = viy * t + (1/2) * a * t^2
20m = (20m/s * sin(18°)) * t + (1/2) * (-9.8m/s²) * t^2
Simplifying:
20m = 20m/s * sin(18°) * t - 4.9m/s² * t^2
Now rearrange the equation:
4.9m/s² * t^2 - 20m/s * sin(18°) * t + 20m = 0
You can solve this quadratic equation using the quadratic formula or by factoring:
t ≈ 1.05 seconds or 3.85 seconds (rounded to two decimal places)
Therefore, the ball takes approximately 1.05 seconds to hit the ground when thrown with an initial velocity of 20m/s at 18 degrees below the horizontal.