Question

ball is thrown horizontally at 7m/s from the top of a hill that 42m high. how far from the base of the cliff would the ball hit the ground

Answer 1

how long does it take to fall 42m? (s = 1/2 at^2)

horizontal speed is constant, so distance = speed * time

Answer 2

h = Vo*T+gT^2 = 42

0+9.8T^2 = 42
T = 2.07s. to hit gnd.
d = V*T = 7*2.07 = 14.49m.

Answer 3

Correction: h = Vo*T+0.5gT^2 = 42

0+4.9T^2 = 42
T = 2.93s.
d = V*T = 7*2.93 = 20.5m.

Answer 4

[AI-generated answer — Explain Bot]

To find how far the ball would hit the ground from the base of the cliff, we can use the kinematic equations of motion.

The given information is:
- Initial velocity of the ball (horizontal component): 7 m/s
- Height of the cliff: 42 m

First, we need to find the time it takes for the ball to hit the ground. Since the ball is thrown horizontally, the initial vertical velocity is 0 m/s. We can use the kinematic equation for vertical motion:

𝑑 = 𝑣₀𝑡 + 1/2𝑎𝑡²

Where:
- 𝑑 is the vertical displacement (−42 m as the ball is falling downwards)
- 𝑣₀ is the initial vertical velocity (0 m/s)
- 𝑎 is the acceleration due to gravity (−9.8 m/s² as it opposes the vertical motion and is downwards)
- 𝑡 is the time

Rearranging the equation gives:

−42 = 0𝑡 + 1/2(−9.8)𝑡²
−42 = −4.9𝑡²

Now, solve for 𝑡 by rearranging the equation:

𝑡² = 42/4.9
𝑡 ≈ √(8.57)
𝑡 ≈ 2.93 s

Next, we need to find the horizontal distance traveled by the ball. The horizontal component of the initial velocity remains constant throughout the motion. We can use the equation:

𝑑 = 𝑣𝑥𝑡

Where:
- 𝑑 is the horizontal distance
- 𝑣𝑥 is the horizontal component of the initial velocity (7 m/s)
- 𝑡 is the time (2.93 s)

Substituting the values:

𝑑 = 7 m/s × 2.93 s
𝑑 ≈ 20.5 m

Therefore, the ball would hit the ground approximately 20.5 meters from the base of the cliff.