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.
I NEED AN ANSWER
26.5
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.
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