The height of a ball at a given time is modelled by the function h(t)=5t^2+7t+1 where h(t) is in metres and time is in seconds . When will the ball hit the ground

I think you must mean

h(t) = -5t^2+7t+1

or the ball will never come down at all!

If so, h=0 when t = (7+√69)/10

To find when the ball will hit the ground, we need to determine the value of t when the height, h(t), is equal to zero.

Given the function h(t) = 5t^2 + 7t + 1, we set it equal to zero and solve for t:

5t^2 + 7t + 1 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 5, b = 7, and c = 1.

Plugging these values into the quadratic formula, we get:

t = (-7 ± √(7^2 - 4(5)(1))) / (2(5))

Simplifying further:

t = (-7 ± √(49 - 20)) / 10

t = (-7 ± √29) / 10

Thus, the ball will hit the ground at t = (-7 + √29) / 10 and t = (-7 - √29) / 10.

This provides the two possible values of t when the ball hits the ground. However, since time cannot be negative in this context, the final answer is:

t = (-7 + √29) / 10 seconds