Question

If a ball is thrown off a 123 m cliff at a speed of 18 m/s, how far from the base of the diff is the ball?

Givens:
Unknown:
Equation:
Substitute:
Solve:

Answer 1

Xo = 18 m/s. = Hor. velocity.

h = o.5g*t^2.
123 = 4.9t^2, t = 5.01 s. to reach gnd.

d = Xo*t = 18 * 5.01 = 90.2 m.

Answer 2

[AI-generated answer — Explain Bot]

To solve this problem, we can use the equation of motion:

𝑑 = 𝑣₀𝑡 + 0.5𝑎𝑡²

In this equation:
- 𝑑 is the distance traveled by the ball,
- 𝑣₀ is the initial velocity of the ball,
- 𝑡 is the time elapsed, and
- 𝑎 is the acceleration due to gravity (which is approximately equal to 9.8 m/s²).

Given:
Initial velocity (𝑣₀) = 18 m/s
Distance (𝑑) = 123 m
Acceleration due to gravity (𝑎) = -9.8 m/s² (negative because acceleration is in the opposite direction of motion)

We need to find the time elapsed (𝑡) when the ball reaches the base of the cliff. To find it, we can rearrange the equation like this:

𝑡 = (−𝑣₀ ± √(𝑣₀² − 4(0.5𝑎)(−𝑑)))/(2(0.5𝑎))

Plugging in the known values:

𝑡 = (−18 ± √(18² − 4(0.5(−9.8))(−123)))/(2(0.5(−9.8)))

Now, simplify the equation and solve for 𝑡:

𝑡 = (−18 ± √(324 + 2352))/−9.8

𝑡 = (−18 ± √(2676))/−9.8

To find the positive value of 𝑡, we ignore the negative value because time cannot be negative in this context.

𝑡 = (−18 + √(2676))/−9.8

Now, substitute the value of 𝑡 into the equation for distance (𝑑):

𝑑 = 𝑣₀𝑡 + 0.5𝑎𝑡²

𝑑 = 18(𝑡) + 0.5(−9.8)(𝑡)²

Substituting the value of 𝑡 into the equation:

𝑑 = 18 * [(−18 + √(2676))/−9.8] + 0.5(−9.8) * [(−18 + √(2676))/−9.8]²

Simplifying the equation will give you the distance (𝑑) from the base of the cliff where the ball lands.