AN OBJECT IS LAUNCHED FROM GROUND LEVEL DIRECTLY UPWARD AT A RATE OF 48 METERS PER SECOND. THE EQUATION FOR THE OBJECTS HEIGHT IS Y =-16ײ+48×=
Use x for a variable name. The × symbol is the multiplication operator.
Use x^2 for x squared.
And did you actually have a question?
I don't mind helping to find answers, but I resent having to provide the questions as well.
and STOP SHOUTING
The equation for the object's height, assuming the initial position is located at the ground level, can be found using the formula for vertical motion:
y = -16t^2 + vt + s
Where:
- y is the height at time t
- v is the initial velocity (in this case, 48 m/s)
- t is the time elapsed
- s is the initial position (in this case, 0 since it starts from ground level)
Plugging in the given values, the equation becomes:
y = -16t^2 + 48t
Therefore, the equation for the object's height is y = -16t^2 + 48t.
It seems that there may be a typo in the equation you provided since there are two equal signs (=) instead of one equal sign. However, I can help you understand how to analyze the motion of the object using the given information.
The equation for the height of an object launched vertically can be represented by the equation:
y = -16t^2 + v₀t + h₀
where:
- y is the height of the object at a given time t,
- t is the time since the object was launched,
- v₀ is the initial velocity (speed) of the object,
- h₀ is the initial height of the object.
In this scenario, the object is launched directly upward from ground level, so the initial height h₀ would be equal to zero.
The given information states that the object is launched with a velocity of 48 meters per second. Therefore, we can substitute v₀ = 48 into the equation:
y = -16t^2 + 48t
Now, to find the height of the object at a specific time t, simply substitute the desired time value into the equation.
For example, if you want to find the height of the object after 2 seconds, substitute t = 2 into the equation:
y = -16(2)^2 + 48(2)
y = -16(4) + 96
y = -64 + 96
y = 32
Therefore, at t = 2 seconds, the object would reach a height of 32 meters.
If the equation provided in your question is incorrect, please check the equation and provide the correct version for a more accurate analysis.