A ball is thrown from a point 1.38 m above the ground. The initial velocity is 19.9 m/s at an angle of 33.8° above the horizontal. Find the maximum height of the ball above the ground.
Calculate the speed of the ball at the highest point in the trajectory.
To find the maximum height of the ball above the ground, we can use the equations of motion for projectile motion.
Step 1: Find the vertical component of the initial velocity.
The vertical component can be calculated using the formula:
Vy = initial velocity * sin(angle)
In this case, the initial velocity is 19.9 m/s and the angle above the horizontal is 33.8°. Therefore, we have:
Vy = 19.9 m/s * sin(33.8°)
Step 2: Calculate the time taken to reach maximum height.
The time taken to reach maximum height can be found using the formula:
t = Vy / g
Here, g represents the acceleration due to gravity (approximately 9.8 m/s²).
Step 3: Find the height at the maximum point.
The height at the maximum point can be found using the formula:
h = initial height + (Vy² / (2 * g))
In this case, the initial height is 1.38 m and we have already calculated Vy and g.
Therefore, we can now substitute the values into the formulas and calculate the maximum height.
Finally, calculate the speed of the ball at the highest point of the trajectory.
To find the speed at the highest point, we can use the formula for the magnitude of velocity:
V = sqrt(Vx² + Vy²)
Since the ball reaches the highest point, its vertical velocity becomes 0. Therefore, we only need to find the horizontal velocity component (Vx) at the highest point.
Given that the horizontal velocity is constant throughout the trajectory (assuming no air resistance), we can use the formula:
Vx = initial velocity * cos(angle)
Substitute the values into the formula to calculate Vx. Finally, using the formula for magnitude of velocity (V), substitute Vx and Vy (0) into the formula and calculate the speed at the highest point.