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