1. A ball is thrown 15 m/s at an angle of 60. How high does the ball go?
2. A ball rolls off a table going 8 m/s. If the table is 1.8 m tall, how far away does the ball land?
1. Vo = 15m/s[60o].
Yo = 15*sin60 = 13 m/s.
Y^2 = Yo^2 + 2g*h.
0 = 13^2 - 19.6h. h = ?.
2. h = 4.9*t^2. 1.8 = 4.9t^2, t = ?.
d = V*t. d = 8*t.
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To solve these problems, we can break them down into two separate components: the vertical motion and the horizontal motion of the objects.
1. To find out how high the ball goes, we need to analyze the vertical motion. We can use the following equation of motion for vertical motion:
π¦ = π¦0 + π£0π¦π‘ - 1/2ππ‘^2
Where:
- π¦ is the vertical position or height
- π¦0 is the initial vertical position or height (in this case, the ground level)
- π£0π¦ is the initial vertical velocity
- π is the acceleration due to gravity (approximately 9.8 m/s^2)
- π‘ is the time taken
Given that the ball is thrown at an angle of 60 degrees with an initial velocity of 15 m/s, we can find the initial vertical velocity (π£0π¦) using trigonometry:
π£0π¦ = π£0 * sin(π)
= 15 m/s * sin(60Β°)
β 12.99 m/s
Now, we can substitute the values into the equation and solve for π¦:
π¦ = 0 + 12.99 m/s * π‘ - 1/2 * 9.8 m/s^2 * π‘^2
To find the height, we need to determine the time (π‘) it takes for the ball to reach its highest point. At the highest point, the vertical velocity will be zero (π£π¦ = 0). Using the equation of motion for vertical velocity:
π£π¦ = π£0π¦ - ππ‘
0 = 12.99 m/s - 9.8 m/s^2 * π‘
Solving for π‘, we find:
π‘ = π£0π¦ / π
= 12.99 m/s / 9.8 m/s^2
β 1.33 s
Now, we can substitute the value of π‘ back into the equation for π¦ to determine the height:
π¦ = 0 + 12.99 m/s * 1.33 s - 1/2 * 9.8 m/s^2 * (1.33 s)^2
β 10 m
Therefore, the ball reaches a height of approximately 10 meters.
2. To find out how far away the ball lands, we need to analyze the horizontal motion. At the instant the ball rolls off the table, it has a horizontal velocity (π£0π₯) that remains constant throughout its flight. This means that the horizontal motion is uniform.
We can use the following equation to calculate the horizontal distance (π₯) traveled by the ball:
π₯ = π£0π₯ * π‘
Given that the ball is rolling off the table with an initial velocity of 8 m/s, we can find the initial horizontal velocity (π£0π₯) using trigonometry:
π£0π₯ = π£0 * cos(π)
= 8 m/s * cos(0Β°)
= 8 m/s
Since the time of flight (π‘) is the same as in the previous question (approximately 1.33 s), we can substitute the values into the equation to calculate π₯:
π₯ = 8 m/s * 1.33 s
β 10.64 m
Therefore, the ball lands approximately 10.64 meters away from the table.