From a platform an object is projected vertically into air. Its height, h metres above ground level after t seconds is modelled by equation h=3+20t-5t^2.

At what height above the ground is the platform from which the object was projected?
Calculate the time when the object will again be at the same height as the platform
At what time, correct to 2 decimal places does the object reach the ground?

a) at that time, t = 0, so

h = 3 + 0 + 0 = 3 m
b) you want the t, when h = 3
3 = 3 + 20t - 5t^2
5t^2 - 20t = 0
t(5t - 20) = 0
t = 0 or t = 4

c) when it hits the ground, h = 0
5t^2 - 20t - 3 = 0

solve, using your favourite method of solving quadratics, reject the negative answer.

To find the height of the platform from which the object was projected, we need to determine the value of h when t = 0.

Substitute t = 0 into the equation h = 3 + 20t - 5t^2:

h = 3 + 20(0) - 5(0)^2
h = 3

Therefore, the height of the platform from which the object was projected is 3 meters above the ground level.

To calculate the time when the object will be at the same height as the platform, we need to set the object's height equal to the height of the platform, which is 3. So we have the equation:

3 + 20t - 5t^2 = 3

Simplifying this equation, we get:

20t - 5t^2 = 0

Factoring out t, we can rewrite the equation as:

t(20 - 5t) = 0

Setting each factor equal to zero gives us two possibilities:

t = 0 or 20 - 5t = 0

Solving the second equation for t, we get:

20 - 5t = 0
5t = 20
t = 4

Therefore, the object will be at the same height as the platform at t = 0 seconds and t = 4 seconds.

To determine the time when the object reaches the ground, we need to find the value of t when h = 0.

Substitute h = 0 into the equation h = 3 + 20t - 5t^2:

0 = 3 + 20t - 5t^2

This is a quadratic equation, so we can solve it by factoring or using the quadratic formula. However, in this case, the quadratic doesn't factor nicely, so let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = -5, b = 20, and c = 3. Substituting these values into the quadratic formula, we have:

t = (-20 ± √(20^2 - 4(-5)(3))) / (2(-5))
t = (-20 ± √(400 + 60)) / (-10)
t = (-20 ± √460) / (-10)

Now we can simplify and calculate the time when the object reaches the ground:

t ≈ (-20 + √460) / (-10) ≈ (-20 + 21.47) / (-10) ≈ 0.147 seconds (rounded to 3 decimal places)

t ≈ (-20 - √460) / (-10) ≈ (-20 - 21.47) / (-10) ≈ 4.147 seconds (rounded to 3 decimal places)

Therefore, the object reaches the ground approximately at t = 0.147 seconds and t = 4.147 seconds (rounded to 3 decimal places).