Question

Hi, can you solve the question for me step by step, I will rate up if the working is correct. I will post the answer together with the question.

Answer:

Question 5 This problem analyzes projectile motion. A netball player throws the ball from a point 4.5 metre horizontally from the net as shown in Figure Q5. The ball is thrown witha speed of 9 ms-1 at an angle of projection of α to the horizontal. Assume the acceleration due to gravity is of magnitude g 10 ms2 Figure Q5 Show that during its flight the horizontal and upward vertical displacementsx and y of the ball from its point of projection satisfy the trajectory equation (a) xtan α--x-sec-α. y (7 marks) (b) If the player throws the ball from the position shown in Figure Q5 with 1 m, find the possible angles of projection of the ball in order that the net should be (6 marks) (c) Analyze the vertical component of the ball's velocity to show that the bal (6 marks) (d) From the trajectory equation, and with the given horizontal distance from the on the ball's trajectory actually reach the net from above, for both possible angles of projection. net, find the maximum achievable height of the net hmar, above the point of ection for different angles of projection α. 6 marks) α-30.960, 71.570 (2dp) For α-30.960, уーー1.2 m/s. For α = 71.570, у In both cases, (b) (c) 7.273 m/s. is negative, so ball reach net from above. -0, gives hmax 2.8 m. Equating d(tan α) (d) d(tana)

Answer 1

s) bau Hime ttsec) . ニー10 acceleration- 一g νου:1 cal clt2 9sin (x) (o) =0 22. rorm 2. χ sin(x) cos(d )

(b) ar dzh = ) m 4- tan (K) = 12 ± 182-180 3,3 乂,-tan-113): 7/.57。 34- 4- Cos(a 7.273 m/s naHV In buth ca

8 1 4- Stan-la) 4-Stan(x)--S 4 - 2 ltan 4.S 12.8m