Help Please, ty A ball is launched at 20 meters per second (m/s) from a 60 meter tall platform. The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 20t + 60, where s is in meters.  What is the height above the ground when the object is launched? How long before the object hits the ground after launch? What is the maximum height of the object?

s = height above ground

s = 60 + 20 t - 4.9 t^2 (standard physics equation on earth)

at t = 0
s = 60 (clearly :)

now when does it hit the bleak earth?
That is when s = 0
4.9 t^2 - 20 t -60 = 0
solve quadratic and use the positive t (the negative t was back before you threw it if you had thrown it from the ground)
t = 6.09 or - 2.01
use t = 6.09

now to do the last part there are two obvious ways to get t at the peak
1. look for vertex of parabola
2. look for halfway between t = -2.01 and t = 6.09
I will do it the hard (11) waay by completing the square
4.9 t^2 - 20 t = -(s-60)
t^2 - 4.08 t = -.204 s + 12.2
t^2 - 4.08 t +2.04^2 = -.204 s +12.2 + 4.16

(t-2.04)^2 = -.204(s-80.2)
so
top at 80.2 meters at t = 2.04 s
===============
quick check on time
should be average of 6.09 and -2.01
=4.08 /2 = 2.04 check

thanks for the answer this answer helped me alot

THANK YOU SO MUCH!!

To find the height above the ground when the object is launched, we can substitute t=0 into the equation s(t) = -4.9t^2 + 20t + 60.

s(0) = -4.9(0)^2 + 20(0) + 60
s(0) = 0 + 0 + 60
s(0) = 60

Therefore, the height above the ground when the object is launched is 60 meters.

To find the time it takes for the object to hit the ground after launch, we need to set s(t) equal to 0 and solve for t.

-4.9t^2 + 20t + 60 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is not straightforward, so we'll use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = -4.9, b = 20, c = 60. Let's substitute these values:

t = (-(20) ± √((20)^2 - 4(-4.9)(60))) / (2(-4.9))

Simplifying further:

t = (-20 ± √(400 - (-1176))) / (-9.8)
t = (-20 ± √(1576)) / (-9.8)
t = (-20 ± 39.745) / (-9.8)

Using a calculator:

t ≈ 0.816 seconds or t ≈ 5.051 seconds

Since time cannot be negative, we only consider the positive value. Therefore, the object takes approximately 5.051 seconds to hit the ground after launch.

The maximum height of the object can be found by looking at the vertex of the quadratic equation. The vertex is given by the formula:

t = -b / (2a)

Substituting the values into the equation, we have:

t = -(20) / (2(-4.9))
t ≈ 2.041 seconds

Now, let's find the height at this time by substituting t = 2.041 seconds into the equation s(t):

s(2.041) = -4.9(2.041)^2 + 20(2.041) + 60

s(2.041) ≈ 60.408 meters

Therefore, the maximum height of the object is approximately 60.408 meters.

To find the height above the ground when the object is launched, we need to evaluate the equation for s(t) at t = 0 seconds.

First, let's substitute t = 0 into the equation s(t) = -4.9t^2 + 20t + 60:

s(0) = -4.9(0)^2 + 20(0) + 60
s(0) = 0 + 0 + 60
s(0) = 60

Therefore, the height above the ground when the object is launched is 60 meters.

Next, to find the time it takes for the object to hit the ground after launch, we need to determine when the height (s) is equal to zero. We can set the equation s(t) = 0 and solve for t.

-4.9t^2 + 20t + 60 = 0

Now, we can solve this equation by factoring, using the quadratic formula, or by completing the square. I'll use the quadratic formula in this explanation.

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b^2 - 4ac))/(2a)

In our case, a = -4.9, b = 20, and c = 60. Substituting these values into the quadratic formula:

t = (-20 ± √(20^2 - 4(-4.9)(60))) / (2(-4.9))

Simplifying:

t = (-20 ± √(400 + 1176)) / (-9.8)
t = (-20 ± √(1576)) / (-9.8)
t = (-20 ± √(4 * 394)) / (-9.8)
t = (-20 ± 2√(394)) / (-9.8)

Now, we have two possible solutions for t. The plus sign (+) in the formula gives the time in seconds before the object reaches the ground, and the minus sign (-) gives the time in seconds after the object passes its highest point and is on its way back down.

Calculating these values will give us the time it takes for the object to hit the ground.