Galileo threw a large rock upward from a tower balcony at an angle of 25.0o above the horizontal with an initial speed of 15 m/s. If the rock hits the ground 3.0 s later, how high above ground was the rock when it left Galileo’s hand?
figure the vertical component of initial veloctiy: 15sin25
then
hf=hi+Viv*t-4.9t^2 hf=0, solve for Hi.
I am wondering how Galileo measured time? Stopwatch? How did he measure intial veloctiy?
To solve this problem, we can use the equations of motion to find the height above the ground when the rock left Galileo's hand.
First, let's break down the initial velocity of the rock into horizontal and vertical components.
The initial speed of 15 m/s can be broken down into horizontal and vertical components using trigonometry.
Vertical component: v_y = v * sin(theta)
Horizontal component: v_x = v * cos(theta)
Where:
v is the initial speed of the rock (15 m/s)
theta is the angle above the horizontal (25.0 degrees)
Using the given values, we can find the vertical component of the initial velocity:
v_y = 15 m/s * sin(25.0 degrees)
v_y ≈ 6.38 m/s
Now that we have the vertical component of the initial velocity, we can determine the maximum height the rock reaches by using the equation of motion:
v_y^2 = v_0y^2 + 2 * a * d
Where:
v_y is the final vertical velocity (0 m/s at the top)
v_0y is the initial vertical velocity (6.38 m/s)
a is the acceleration (due to gravity, which is -9.8 m/s^2)
d is the height above the ground.
Rearranging the equation, we can solve for d:
d = (v_y^2 - v_0y^2) / (2 * a)
Substituting the values, we get:
d = (0 - (6.38 m/s)^2) / (2 * -9.8 m/s^2)
d ≈ 1.67 meters
Therefore, the rock was approximately 1.67 meters above the ground when it left Galileo's hand.