A ball is fired from the ground with an initial speed of 1.36 103 m/s (approximately four times the speed of sound) at an initial angle of 49.0° to the horizontal. Neglect air resistance.
You have neglected to say what quantity you want to be computed. The horizontal distance that it travels?
That would be (Vo^2/g)*sin98 = 18690 m
(18.69 km)
swag
To solve this problem, we can break it down into two components: the horizontal and vertical motions of the ball.
1. Horizontal Motion:
Since there is no acceleration in the horizontal direction, the velocity of the ball remains constant throughout its motion. The initial horizontal velocity can be found using the initial speed and the angle of projection:
vx = initial speed × cos(angle)
= 1.36 x 10^3 m/s × cos(49.0°)
≈ 877.6 m/s
2. Vertical Motion:
The vertical motion of the ball is influenced by gravity, so we need to consider the equations of motion in the vertical direction. We can use the following equations:
- Vertical velocity component at any given time (vy) can be found using the following formula:
vy = initial speed × sin(angle)
= 1.36 x 10^3 m/s × sin(49.0°)
≈ 9.59 x 10^2 m/s
- The time it takes for the ball to reach the highest point (t_top) can be found using the formula:
t_top = vy / g, where g is the acceleration due to gravity (9.8 m/s²).
= (9.59 x 10^2 m/s) / (9.8 m/s²)
≈ 97.9 seconds
- The time it takes for the ball to return to the ground can be found by using the formula:
t_total = 2t_top
= 2 x 97.9
≈ 195.8 seconds
It's important to note that neglecting air resistance assumes a vacuum-like condition, which is not realistic for most situations on Earth.