Question

A soccer ball is kicked horizontally off a 22.0 meter high hill and lands a distance of 35 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

Answer 1

If this is an honors class, you clearly need to do some analysis.

What time does an object take to fall 22m?

in that time, the ball travels 35 m.

Velocity=distance/time

Answer 2

I NEED AN ANSWER

Answer 3

26.5

Answer 4

[AI-generated answer — Explain Bot]

To determine the initial horizontal velocity of the soccer ball, we can use the kinematic equation for horizontally launched projectiles:

d = vₓ * t

where:
d is the horizontal distance traveled (35 meters),
vₓ is the initial horizontal velocity, and
t is the time of flight.

In this case, we need to first find the time of flight. To do that, we can use the kinematic equation for vertical motion:

h = v₀ᵥ * t + (1/2) * g * t²

where:
h is the vertical height (22.0 meters),
v₀ᵥ is the initial vertical velocity, and
g is the acceleration due to gravity (-9.8 m/s²).

Since the ball is launched horizontally, there is no initial vertical velocity (v₀ᵥ = 0).

Substituting the known values into the equation:

22.0 = (1/2) * (-9.8) * t²

Simplifying the equation:

t² = -4.49

Since time cannot be negative, we discard the negative solution. Therefore:

t ≈ √4.49 ≈ 2.12 seconds

Now that we have determined the time of flight, we can substitute that value back into the horizontal distance equation:

35 = vₓ * 2.12

Solving for vₓ:

vₓ ≈ 35 / 2.12 ≈ 16.51 m/s

Thus, the initial horizontal velocity of the soccer ball is approximately 16.51 m/s.