Question

Assume that an object is launched upward at 980 m/sec. Its position would be given by s = -4.9t² + 980t. Find the maximum height attained by the object.

Answer 1

Answer:

Step-by-step explanation:

Max height=u²/2g

980²/2g

49000m is your ans

Use the third eqn

Answer 2

The maximum height attained by the object is 49000 meters.

Step-by-step explanation:

The given equation is

$s=-4.9t^2+980t$

Comparing this equation with $y=ax^2+bx+c$

a = -4.9, b = 980, c = 0

Since, a<0 . Hence it represents a downward parabola.

And for a downward parabola, the vertex is the highest point. hence, the maximum value is at vertex.

x- coordinate of vertex is given by

$x=-\frac{b}{2a}\\\\=-\frac{980}{2\cdot(-4.9)}\\\\=100$

y- coordinate of vertex is given by

$y=s(100)\\\\=-4.9(100)^2+980(100)\\\\=-49000+98000\\\\=49000$

Therefore, the maximum height attained by the object is 49000 meters.

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