Bill is standing on a bridge and kicks stones to the water below. If bill kicks a stone with a horizontal velocity of 3.5 m/s, and it lands in the water at a horizontal distance of 5.4m. What is the height of the bridge?
Do I need to find the time first?? If so can I use a rewritten version of Vf = Vi + gt where Vf = 0?
Time, t = horiz. distance / horiz. velocity
Solve for vertical distance, H using
H = v0t+(1/2)gt²
To find the height of the bridge, you need to calculate the time it takes for the stone to land in the water. Then, you can use this time to determine the height using the equation of motion.
First, let's find the time it takes for the stone to land. Since the horizontal velocity remains constant (no horizontal acceleration), you can use the equation:
Distance = Speed × Time
Substituting the values you have, where distance = 5.4m and speed = 3.5 m/s:
5.4m = 3.5 m/s × Time
Solving for Time:
Time = 5.4m / 3.5 m/s ≈ 1.54 seconds
Now that you have the time (approximately 1.54 seconds), you can calculate the height of the bridge using the equation of motion for vertical motion. Since the vertical motion is influenced by gravity, you can use the equation:
Vertical Distance = Initial Vertical Velocity × Time + (0.5) × Acceleration × Time²
Where Initial Vertical Velocity is the vertical velocity when the stone is kicked and Acceleration is the acceleration due to gravity (approximately 9.8 m/s²).
In this case, the stone is kicked horizontally, so the Initial Vertical Velocity is 0 m/s. We can ignore the first term of the equation, and the equation reduces to:
Vertical Distance = (0.5) × Acceleration × Time²
Substituting the values, we have:
Vertical Distance = (0.5) × 9.8 m/s² × (1.54 seconds)²
Calculating this, we get:
Vertical Distance ≈ 1.50 meters
Therefore, the height of the bridge is approximately 1.50 meters.