H(t)=-4.9^2+280+7

Question

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Answer 1

h(t) = -4.9t^2 + 280t + 7 seems to be the corrected form of the equation, as t^2 and t were accidentally miscopied or omitted

h(t) = height in meters after t seconds with initial velocity = 280 meters per second, and initial height = 7 meters

general equation is

h(t) = (a/2)t^2 + vot + ho, where vo = initial velocity, ho = initial height, a = acceleration due to gravity

velocity at time t = v(t)=the derivative of h(t) = h'(t) = -9.8t + 280

set it equal to zero to find time of maximum height

-9.8t +280 = 0

t = 280/9.8 = 140/4.9 = 20/.7 = 200/7 = 28 4/7 seconds

plug that value into h(t) to find the maximum height

h(200/7) = -4.9(200/7)^2 + 280(200/7) + 7

to find when the object hits the ground, set h(t) = 0

and solve for t

-4.9t^2 + 240t + 7 = 0

use the quadratic formula

t = -240/-9.8 + (1/9.8)sqr(240^2 +4(4.9)(7))

those are the usual type of questions that come up with that h(t) equation

it looks like the equation was miscopied leaving out the t^2 and t

the -4.9 is the clue that the height is measured in meters

for feet it would have been -16, the t^2 term is the effect of gravity

the force of gravity at sea level on earth is about 9.8 meters per second per second

= about 32 feet per second per second

two other ways to solve or view the equation is graph it. it's a downward opening parabola. the max value is the vertex of the parabola. Use a graphing calculator, on line or hand held.

another way is rewrite the h(t) polynomial in vertex form. the vertex (t,h) will mean t= time of max height and h= max height

h(t) = -4.9t^2 + 280t + 7

= -4.9(t^2 - (280/4.9)t) +7

= -4.9(t - 140/4.9)^2 + 7 + (140/4.9)^2

the vertex is (140/4.9, 7+ (140/4.9)^2)

where the x coordinate is the time t of max height

and the y coordinate is the max height, h.

since this problem is listed under the topic "algebra 2" odds are you want the last method. the 1st method involves differential calculus, a more advanced category of mathematics

but if you really meant to write H(t) = -4.9^2 +280+7

then

H(t) = -(4.9^2) +287

H(t) = -24.01 +287

H(t) = 262.99

another possibility or perspective is you used H(t) as an integral of h(t)

similar to how F(x) is the integral or antiderivative of f(x)

then

H(t) =-4.9t^2 +280t +7

take the derivative

h(t) =-9.8t +280

which all shows up in the first above calculations with h(t) = v(t) = velocity at time t

or if H(t) =262.99

then h(t) = H'(t) = 0