an object is thrown upwards and its height h(t) in meters after t seconds is given by the equation h(t)=1+4t-5t^2

.when the object hit the ground.

0.4sec

To find when the object hits the ground, we need to determine the value of t when h(t) is equal to zero.

Given the equation h(t) = 1 + 4t - 5t^2, we set it equal to zero:

0 = 1 + 4t - 5t^2

Rearrange the equation to match the quadratic form:

5t^2 - 4t - 1 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 5, b = -4, and c = -1.

Substituting these values into the quadratic formula, we get:

t = (-(-4) ± √((-4)^2 - 4(5)(-1))) / (2(5))

Simplifying further,

t = (4 ± √(16 + 20)) / 10
t = (4 ± √36) / 10
t = (4 ± 6) / 10

This gives us two possible values for t:

1. When t = (4 + 6) / 10 = 10 / 10 = 1
2. When t = (4 - 6) / 10 = -2 / 10 = -0.2

Since time cannot be negative in this context, the object hits the ground approximately 1 second after it is thrown upwards.

To find when the object hits the ground, we need to determine the value of t when the height h(t) is equal to zero. In other words, we need to solve the equation h(t) = 0.

Given the equation for the height of the object as h(t) = 1 + 4t - 5t^2, we set it equal to zero and solve for t:

0 = 1 + 4t - 5t^2

Rearranging the equation, we get:

5t^2 - 4t - 1 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = 5, b = -4, and c = -1. Plugging these values into the quadratic formula, we get:

t = (-(-4) ± √((-4)^2 - 4(5)(-1))) / (2(5))
t = (4 ± √(16 + 20)) / 10
t = (4 ± √36) / 10
t = (4 ± 6) / 10

This gives us two possible solutions:

t1 = (4 + 6) / 10 = 10 / 10 = 1
t2 = (4 - 6) / 10 = -2 / 10 = -0.2

Since time cannot be negative in this case, the object hits the ground after 1 second.

its height is zero when it hits the ground, right? So, just solve

1+4t-5t^2 = 0