Projectile problems

Question: I am currently still in yr 11, but have not long till starting yr 12. I have started doing projectile motions in tutoring college that I go to.

A baseball is hit with a velocity of 28.0 m/s at an angle of 30.0^o to the horizontal at an initial height of 1.00 m above the plate.

(a) What is the magnitude of the vertical component of the initial velocity?
(b) How long does it take the ball to return to the initial height above the ground?
(c) What is the horizontal distance the ball travels before returning to the initial height above the ground?
(d) The ball is hit directly towards a stationary outfielder who is 85.0 m from the plate.
At the instant the ball is hit, she begins to run towards the plate with constant acceleration.
What is the magnitude of her acceleration if she is to catch the ball when it is 1.00 m above the ground?

Answer:

(a) Vertical component = 28sin(30) = 14 m/s

(b) Using v = u + at with v = -14 m/s, u = 14 m/s and a = g = -9.8 m/s² (minus because it acts downwards)
-28 = - 9.8t therefore t = 28/9.8 = 2.86 s.

(c) Horizontal component = 28cos(30) = 24.25 m/s
Horizontal distance travelled in 2.86 s = 24.25x2.86 = 69.35 m

(d) It takes 2.86 s for the ball to return to 1 m above the ground and it will have travelled 69.35.
Therefore the fielder will have had to run forward a distance of 85 – 69.35 = 15.65 m.
Initail velocity = 0, distance = 15.65 m, time = 2.86.
Use: s = ut + ½ at²

15.65 = 0 + ½ a(2.86x2.86) so acceleration (a) = 2x15.65/8.18 = 3.83 m/s².