A ball is thrown from a point 1.1 m above the ground. The initial velocity is 20.3 m/s at an angle of 32.0° above the horizontal.
(a) Find the maximum height of the ball above the ground???
(b) Calculate the speed of the ball at the highest point in the trajectory???
(a) Maximum height H is where
g*H = (Vo*sin32)^2/2
(b) Vo*cos35, which remains constant while the vertical velocity component becomes zero at max height.
What would the Vo be???
0 or 20.3???
To find the maximum height of the ball above the ground (a), we need to analyze the vertical motion of the ball. We can break the initial velocity into its vertical and horizontal components. The vertical component can be found using trigonometry:
Vertical velocity (Vy) = Initial velocity (V) * sin(angle)
Vy = 20.3 m/s * sin(32.0°)
Next, we can use physics equations to determine the maximum height. At the maximum height, the vertical velocity becomes 0, so we can use the formula:
Vy^2 = V^2 - 2 * g * Δy
Where:
Vy = Final vertical velocity
V = Initial velocity
g = Acceleration due to gravity (approximately 9.8 m/s^2)
Δy = Vertical displacement
Rearranging the formula:
Δy = (Vy^2) / (2 * g)
Now let's substitute the values:
Δy = (0 - (20.3 m/s * sin(32.0°))^2) / (2 * 9.8 m/s^2)
Simplifying:
Δy = - (20.3 m/s * sin(32.0°))^2 / (2 * 9.8 m/s^2)
So, the maximum height of the ball above the ground is the absolute value of Δy.