Can someone please answer this question please answer it correctly and please show work please help me I need it
2 answers:
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Answer:
- h = -16t^2 + 73t + 5
- h = -16t^2 + 5
- h = -4.9t^2 + 73t + 1.5
- h = -4.9t^2 + 1.5
Step-by-step explanation:
The general equation we use for ballistic motion is ...
where g is the acceleration due to gravity, v₀ is the initial upward velocity, and h₀ is the initial height.
The values of g commonly used are -32 ft/s², or -4.9 m/s². Units are consistent when the former is used with velocity in ft/s and height in feet. The latter is used when velocity is in m/s, and height is in meters.
_____
Dwayne throws a ball with an initial velocity of 73 feet/second. Dwayne holds the ball 5 feet off the ground before throwing it. (h = -16t^2 + 73t + 5)
A watermelon falls from a height of 5 feet to splatter on the ground below. (h = -16t^2 + 5)
Marcella shoots a foam dart at a target. She holds the dart gun 1.5 meters off the ground before firing. The dart leaves the gun traveling 73 meters/second. (h = -4.9t^2 + 73t + 1.5)
Greg drops a life raft off the side of a boat 1.5 meters above the water. (h = -4.9t^2 + 1.5)
_____
<em>Additional comment on these scenarios</em>
The dart and ball are described as being launched at 73 units per second. Generally, we expect launches of these kinds of objects to have a significant horizontal component. However, these equations are only for <em>vertical</em> motion, so we must assume the launches are <em>straight up</em> (or that the up-directed component of motion is 73 units/second).
Answer:
the positive slope of the asymptote = 5
Step-by-step explanation:
Given that:
Using the standard form of the equation:
where:
(h,k) are the center of the hyperbola.
and the y term is in front of the x term indicating that the hyperbola opens up and down.
a = distance that indicates how far above and below of the center the vertices of the hyperbola are.
For the above standard equation; the equation for the asymptote is:
where;
is the slope
From above;
(h,k) = 11, 100
= 100
a =
a = 10
b =
b = 2
y = 5x-53 , -5x -57
Since we are to find the positive slope of the asymptote: we have
to be the slope in the equation
=
= 5
Thus, the positive slope of the asymptote = 5