an object is fired upward at the end of the burn it has an upward velocity of 245m/s and is 14.7 m high

a) find the maximum height and when it is attained
b) when it reaches the ground

i know that i have to use the quadratic function s= -4.9t²+Vot+h

but for what i have the 4.9t² how can i use it and can help me with the problem please

For the max height, energy thinking

top PEnergy = intial PE + Initial Kenergy
mgh=mg*14.7 + 1/2 m 245^2
solve for h, the maximum height.

when is it attained?
h=14.5+245t -1/2 g t^2 for h max, solve for time t.
when it reaches the ground?
h=0 solve for time.

The 4.9 is a fixed constant and deals with the force of gravity on earth.

Your equation would be
s = -4.9t^2 + 245t + 14.7

take the derivative set it equal to zero and solve for t
put that t back into the equation to find the maximum height

b) when it hits the ground, s = 0
so solve 0 = -4.9t^2 + 245t + 14.7 for t

great job

To solve this problem, you can use the equation of motion for an object under constant acceleration:

s = -4.9t² + V₀t + h

Where:
s is the height (vertical position) of the object,
t is the time,
V₀ is the initial velocity of the object (the velocity at the end of the burn, which is 245 m/s in this case),
h is the initial height of the object (which is 14.7 m in this case),
and -4.9 is a constant representing acceleration due to gravity (it's negative because it acts downward).

a) To find the maximum height and when it is attained, we need to find the vertex of the quadratic equation. The vertex represents the highest point of the object's trajectory.

The vertex of a quadratic equation of the form ax² + bx + c is given by the formula: t = -b / (2a).

In this case, the equation is -4.9t² + 245t + 14.7. To find the time of maximum height, replace a with -4.9 and b with 245.

t = -245 / (2 * -4.9)
t = -245 / -9.8
t = 25 seconds

Now, substitute this value of t back into the equation to find the maximum height:

s = -4.9(25)² + 245(25) + 14.7
s = -4.9(625) + 6125 + 14.7
s = -3062.5 + 6125 + 14.7
s = 3062.5 + 6125 + 14.7
s = 9202.2 meters

Therefore, the maximum height is 9202.2 meters, and it is attained after 25 seconds.

b) To find when the object reaches the ground, we need to find the time at which the height becomes zero.

Set s = 0 and solve for t:

0 = -4.9t² + 245t + 14.7

This equation is a quadratic equation which can be solved by factoring, using the quadratic formula, or graphical methods. However, since this equation does not factor nicely, let's use the quadratic formula to find the time when the object reaches the ground:

t = (-b ± √(b² - 4ac)) / (2a)

Substituting the values from the equation:

t = [ -245 ± √(245² - 4(-4.9)(14.7)) ] / (2 * -4.9)

t ≈ [ -245 ± √(60025 + 287.28) ] / -9.8

t ≈ [ -245 ± √(60312.28) ] / -9.8

Since time cannot be negative, we only consider the positive solution:

t ≈ [ -245 + √(60312.28) ] / -9.8

t ≈ 25.09 seconds

Therefore, the object reaches the ground after approximately 25.09 seconds.