a grapefruit is tossed straight up with an initial velocity of 50 ft/sec. the grapefruit is 5ft above ground when released. how high does it go before returning to the ground equation? Equation of height in terms of time y= -16t^2+50t+5
y = -16(t - 25/16)^2 + 705/16
so, the vertex is at (25/16,705/16)
That should help.
CHUSS
The equation you provided for the height of the grapefruit in terms of time is correct. The equation is:
y = -16t^2 + 50t + 5
Where:
y = height of the grapefruit above the ground (in feet)
t = time in seconds
To find how high the grapefruit goes before returning to the ground, we need to find the time when the height is 0.
Set y = 0 in the equation:
0 = -16t^2 + 50t + 5
Now we have a quadratic equation. We can solve it using various methods, such as factoring, completing the square, or using the quadratic formula.
In this case, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -16, b = 50, and c = 5. Substituting these values into the formula:
t = (-50 ± √(50^2 - 4(-16)(5))) / (2(-16))
Simplifying the equation:
t = (-50 ± √(2500 + 320)) / (-32)
t = (-50 ± √2820) / -32
t = (-50 ± 53.09) / -32
There are two possible solutions for time, but since we are interested in the time it takes for the grapefruit to reach the ground, we only consider the positive solution:
t = (-50 + 53.09) / -32
t = 3.09 / -32
t ≈ -0.097 seconds
Since time cannot be negative in this case, we discard the negative solution.
Therefore, the grapefruit will take approximately 0.097 seconds to reach the ground.
Now, to find the maximum height reached by the grapefruit, substitute this value back into the equation:
y = -16(0.097)^2 + 50(0.097) + 5
y ≈ -0.15 + 4.85 + 5
y ≈ 9.7 feet
So, the grapefruit will reach a maximum height of approximately 9.7 feet before returning to the ground.
To determine how high the grapefruit goes before returning to the ground, we can use the equation of height in terms of time, which is given as:
y = -16t^2 + 50t + 5
In this equation, "y" represents the height of the grapefruit at any given time "t". The term "t^2" is the square of the time, and the coefficients "-16" and "50" describe the effects of gravity and initial velocity, respectively. The constant term "+5" represents the initial height of the grapefruit.
To find the highest point reached by the grapefruit, we need to find the vertex of the parabolic function, as the vertex represents the maximum or minimum point of the parabola.
The vertex of a parabolic function in the form "y = ax^2 + bx + c" can be calculated using the formula:
t_vertex = -b / (2a)
In our case, a = -16 and b = 50, so we can substitute these values into the formula:
t_vertex = -50 / (2 * -16)
= -50 / -32
= 1.5625 seconds
Now that we know the time at which the grapefruit reaches its highest point (t_vertex = 1.5625 seconds), we can substitute this value back into the original equation to find the corresponding height:
y = -16(1.5625)^2 + 50(1.5625) + 5
Simplifying this equation will give us the height reached by the grapefruit before returning to the ground.