A ball is thrown from a point 1.2 m above the ground. The initial velocity is 19.8 m/s at an angle of 34.0° 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.
Break the velocty into vertical and horizontal components.
Then, height=1.2 + vivertical*t-4.9t^2
and hmax occurs at the t when
Vv=0=vivertical -9.8t
To find the maximum height of the ball above the ground, we can use the principles of projectile motion. We'll need to use the equations of motion in both the horizontal and vertical directions.
(a) Find the maximum height of the ball above the ground:
1. We need to find the time it takes for the ball to reach its maximum height. To do this, we can use the vertical component of the initial velocity (19.8 m/s) and the acceleration due to gravity (-9.8 m/s^2).
Use the equation: v = u + at, where:
v = final velocity (0 m/s at the highest point)
u = initial velocity in the vertical direction (19.8 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time taken to reach the maximum height (what we need to find)
Rearranging the equation, we get:
0 = 19.8 - 9.8t
Solve for t:
9.8t = 19.8
t = 19.8 / 9.8
t ≈ 2.02 seconds
2. Now, we can determine the maximum height by finding the displacement in the vertical direction during this time.
Use the equation: s = ut + (1/2)at^2, where:
s = displacement in the vertical direction (what we need to find)
u = initial velocity in the vertical direction (19.8 m/s)
t = time taken to reach the maximum height (2.02 seconds)
a = acceleration due to gravity (-9.8 m/s^2)
Plugging in the values:
s = (19.8)(2.02) + (1/2)(-9.8)(2.02)^2
Calculate the result:
s ≈ 19.92 meters
Therefore, the maximum height of the ball above the ground is approximately 19.92 meters.
(b) Calculate the speed of the ball at the highest point in the trajectory:
At the highest point of the trajectory, the vertical component of the velocity is zero, but the horizontal component remains constant.
1. To find the horizontal component of the velocity, given the initial velocity (19.8 m/s) and the launch angle (34.0°), we can use trigonometry.
Use the equation: Vx = V * cos(angle), where:
Vx = horizontal component of the velocity (what we need to find)
V = initial velocity (19.8 m/s)
angle = launch angle (34.0°)
Plugging in the values:
Vx = 19.8 * cos(34.0°)
Calculate the result:
Vx ≈ 16.42 m/s
2. The speed at the highest point in the trajectory is equal to the magnitude of the velocity vector, which is the vector sum of the horizontal and vertical components.
Use the equation: speed = sqrt(Vx^2 + Vy^2), where:
speed = speed at the highest point (what we need to find)
Vx = horizontal component of the velocity (16.42 m/s)
Vy = vertical component of the velocity (0 m/s)
Plugging in the values:
speed = sqrt(16.42^2 + 0^2)
Calculate the result:
speed ≈ 16.42 m/s
Therefore, the speed of the ball at the highest point in the trajectory is approximately 16.42 m/s.