A ball thrown into the air lands 38 m away 2.4 s later. Find the direction and magnitude of the initial velocity.
To find the direction and magnitude of the initial velocity of the ball, we can use the equations of motion.
First, let's break down the problem into its components. We'll consider the horizontal (x) and vertical (y) directions separately.
In the horizontal direction (x-direction), we know that the ball travels 38 m in 2.4 s. Since there is no acceleration in the x-direction (assuming no air resistance or horizontal forces), the velocity in the x-direction remains constant throughout the motion. Therefore, we can use the formula:
distance = velocity x time
38 m = (velocity in the x-direction) x (2.4 s)
Solving for the velocity in the x-direction:
(velocity in the x-direction) = 38 m / 2.4 s
(velocity in the x-direction) = 15.83 m/s
Since there is no force acting in the x-direction, the velocity remains constant throughout the motion in that direction. Therefore, the velocity in the x-direction is the initial horizontal velocity.
Now let's move on to the vertical direction (y-direction). We need to find the initial vertical velocity component of the ball. We know that the ball takes 2.4 seconds to reach the ground again, so the total time of the upward and downward journey is 2.4 seconds divided by 2, which is 1.2 seconds.
In the vertical direction, we can use the formula:
v = u + at
Where:
v = final velocity in the y-direction
u = initial velocity in the y-direction (which we need to find)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
At the highest point, the velocity is zero since the ball momentarily stops before starting to fall back. Therefore, we can use this information to find the initial velocity in the y-direction.
0 = (initial velocity in the y-direction) + (-9.8 m/s^2) x (1.2 s)
Solving for the initial velocity in the y-direction:
(initial velocity in the y-direction) = 9.8 m/s^2 x 1.2 s
(initial velocity in the y-direction) = 11.76 m/s
Now we have the initial velocities in both the x and y directions. To find the magnitude and direction of the initial velocity, we can use the Pythagorean theorem:
magnitude of the initial velocity = sqrt((velocity in the x-direction)^2 + (velocity in the y-direction)^2)
magnitude of the initial velocity = sqrt((15.83 m/s)^2 + (11.76 m/s)^2)
magnitude of the initial velocity = sqrt(250.9561 m^2/s^2 + 138.2976 m^2/s^2)
magnitude of the initial velocity = sqrt(389.2537 m^2/s^2)
magnitude of the initial velocity ≈ 19.73 m/s
To find the direction, we can use the formula for the angle:
angle = atan((velocity in the y-direction) / (velocity in the x-direction))
angle = atan(11.76 m/s / 15.83 m/s)
angle ≈ 36.6 degrees
Therefore, the magnitude of the initial velocity is approximately 19.73 m/s, and the direction of the initial velocity is approximately 36.6 degrees above the horizontal direction.