In a hockey match a hockey ball is hit 18m from the front of the goal. The ball leaves the hockey stick with initial velocity v at angle theta to the horizontal ground. The ball passes over the goal at a maximum height of 2.0m.

The initial vertical component of the velocity of the ball is 6.3ms-1. Air resistance has negligible effect on the motion of the ball.
Show that the time take for t to reach maximum height is about 0.6s.

I used suvat and got
s-2
u-6.3
no v
a-9.81
t ??????
so s=ut + 1/2at^2 has to be used to find t. I kept trying to rearrange find the answer but couldn’t do please help using suvat!!!

To find the time taken for the ball to reach its maximum height, we can use the equation s = ut + (1/2)at^2, where:

- s is the displacement (maximum height) of the ball,
- u is the initial vertical component of the velocity (6.3 m/s),
- a is the acceleration due to gravity (-9.81 m/s^2), and
- t is the time taken to reach the maximum height.

Since we know that the maximum height is 2.0 m, we can substitute these values into the equation to solve for t:

2.0 = 6.3t + (1/2)(-9.81)t^2

To simplify the equation, we can multiply every term by 2 to eliminate the fraction:

4.0 = 12.6t - 9.81t^2

Now, rearrange the equation to get it in quadratic form:

9.81t^2 - 12.6t + 4.0 = 0

Now we have a quadratic equation in the form at^2 + bt + c = 0, where:
- a = 9.81,
- b = -12.6,
- c = 4.0

To solve for t, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values into the quadratic formula, we get:

t = (-(-12.6) ± √((-12.6)^2 - 4(9.81)(4.0))) / (2(9.81))

t = (12.6 ± √(158.76 - 156.84)) / 19.62

t = (12.6 ± √1.92) / 19.62

t = (12.6 ± 1.39) / 19.62

Now, we have two possible values for t:

1) t = (12.6 + 1.39) / 19.62 = 13.99 / 19.62 ≈ 0.713 s

2) t = (12.6 - 1.39) / 19.62 = 11.21 / 19.62 ≈ 0.570 s

Since time cannot be negative, the time taken for the ball to reach its maximum height is approximately 0.570 s.