a rock is thrown 1.8 meters into the air. find how fast it was thrown?
To find how fast the rock was thrown, we can use the equation of motion relating displacement, initial velocity, time, and acceleration.
The equation is:
displacement = initial velocity * time + 0.5 * acceleration * time^2
In this case, the displacement is 1.8 meters. Since the rock is thrown vertically upwards, the acceleration due to gravity (-9.8 m/s^2) will act in the opposite direction. The acceleration will be negative.
We have:
1.8 = initial velocity * time - 0.5 * 9.8 * time^2
Since we need to find the initial velocity, we can eliminate the time variable by using another equation.
When the rock reaches its peak height, its final velocity will be zero. So we can use another equation:
final_velocity = initial_velocity + acceleration * time
At the highest point, the final velocity is zero. Therefore:
0 = initial_velocity - 9.8 * time
We can solve this equation for time:
time = initial_velocity / 9.8
Substituting this time value into the first equation:
1.8 = initial_velocity * (initial_velocity / 9.8) - 0.5 * 9.8 * (initial_velocity / 9.8)^2
Simplifying and rearranging the equation:
1.8 = initial_velocity^2 / 9.8 - 0.5 * initial_velocity^2 / 9.8
Combining like terms:
1.8 = (initial_velocity^2 - 0.5 * initial_velocity^2) / 9.8
Multiplying both sides by 9.8 to eliminate the denominator:
17.64 = (initial_velocity^2 - 0.5 * initial_velocity^2)
Combining like terms:
17.64 = (0.5 * initial_velocity^2)
Dividing both sides by 0.5:
35.28 = initial_velocity^2
Taking the square root of both sides:
initial_velocity = √(35.28)
Calculating the square root:
initial_velocity ≈ 5.94 m/s
Therefore, the rock was thrown with an approximate initial velocity of 5.94 meters per second.