y=-16t2nd power+50x
max hgiht:
total time in air:
starting launch point:
height of 2 seconds:
Hight of 5.3 seconds:
+23254 Because the problem isn't stated clearly, I'm going to assume that it was tossed from the ground, that the time is in seconds, and that the distance is in feet.
If you graphed this equation, you would get a parabola. Also, if you graphed it, you would get the vertex, which helps answer some of the questions.
We need to know the vertex; the x-value of the vertex is found by using the expression: -b/(2a).
Since the formula is: y = -16t² + 50t, a = -16 and b = 50
The x-value at the vertex is: -50/(2·-16) = -50/-32 = 1.5625.
1.5625 seconds is the time that it takes to get from ground level to the highest point; it will take another 1.5625 seconds to fall back to the ground. This results in a total launch times of 2(1.5625) or 3.125 seconds.
At 1.5625 seconds, the object will be at its maximum height; put in this value into the equation:
y = -16(1.4625)² + 50(1.5625) = 39.0625 feet <--- maximum height
To find its height at 2 seconds: y = -16(2)² + 50(2) = 36 feet
Since it hits the ground at 3.125 seconds, at 5.3 seconds, it's laying on the ground (?).
+23254 Because the problem isn't stated clearly, I'm going to assume that it was tossed from the ground, that the time is in seconds, and that the distance is in feet.
If you graphed this equation, you would get a parabola. Also, if you graphed it, you would get the vertex, which helps answer some of the questions.
We need to know the vertex; the x-value of the vertex is found by using the expression: -b/(2a).
Since the formula is: y = -16t² + 50t, a = -16 and b = 50
The x-value at the vertex is: -50/(2·-16) = -50/-32 = 1.5625.
1.5625 seconds is the time that it takes to get from ground level to the highest point; it will take another 1.5625 seconds to fall back to the ground. This results in a total launch times of 2(1.5625) or 3.125 seconds.
At 1.5625 seconds, the object will be at its maximum height; put in this value into the equation:
y = -16(1.4625)² + 50(1.5625) = 39.0625 feet <--- maximum height
To find its height at 2 seconds: y = -16(2)² + 50(2) = 36 feet
Since it hits the ground at 3.125 seconds, at 5.3 seconds, it's laying on the ground (?).
