The vertical initial Velocity component is
Vyo = 10 sin60 = 8.66 m/s
in either coordinate system.
The ball will rise a distance H given by
g H = (1/2) Vyo^2 = 3.82 m
Vyo = 10 sin60 = 8.66 m/s
in either coordinate system.
The ball will rise a distance H given by
g H = (1/2) Vyo^2 = 3.82 m
First, let's break down the initial velocity of the ball into its vertical and horizontal components. The initial velocity can be split into two parts: the horizontal velocity (along the track) and the vertical velocity.
Given:
Initial velocity = 10.0 m/s
Launch angle = 60.0 degrees
To find the vertical component of the initial velocity, we can use the formula:
Vertical velocity (Vy) = Initial velocity * sin(angle)
Vertical velocity (Vy) = 10.0 m/s * sin(60.0 degrees)
Vy ā 8.66 m/s
Now, we know that the ball will rise and then fall back down due to gravity. At the highest point, the vertical velocity will be zero. We can use this information to find the time it takes for the ball to reach its highest point.
Using the equation for vertical motion:
Vertical velocity (Vf) = Vy + (acceleration * time)
Since the ball reaches its highest point, Vf (final vertical velocity) is zero:
0 m/s = 8.66 m/s + (-9.8 m/s^2 * time)
Solving for time:
time = -8.66 m/s / -9.8 m/s^2
time ā 0.884 seconds
Now, we can use this time to find out how high the ball rises. Using the equation for vertical displacement:
Vertical displacement = (initial vertical velocity * time) + (0.5 * acceleration * time^2)
Since the initial vertical velocity is 8.66 m/s and the acceleration due to gravity is -9.8 m/s^2 (negative due to the direction), we have:
Vertical displacement = (8.66 m/s * 0.884 seconds) + (0.5 * -9.8 m/s^2 * (0.884 seconds)^2)
Vertical displacement ā 3.83 meters
Therefore, the professor sees the ball rise approximately 3.83 meters high.