Question

(c) The ball is at its maximum height after _____ seconds.

(d) The maximum height of the ball is ______ meters

(e) The ball traveled a total horizontal distance of _______ meters.

(ROUND TO TWO DECIMAL PLACES)

A baseball player throws a baseball with an initial speed of 155 feet per second at an angle of 23 to the horizontal. The ball leaves his hand at a height of 6 feet. (a) Find parametric equations that describe the position of the ball as a function of time. (b) How long is the ball in the air? (c) When is the ball at its maximum height? Determine the maximum height of the ball. (d) Determine the horizontal distance that the ball traveled. (0) Using a graphing utility, simultaneously graph the equations found in part (a). Assume g=32 ft/sec/sec. (a) Choose the correct equations below. O A. x= (6 cos 23), y=-161? + (6 sin 23*)t + 155 O B. X= (23 cos 6°)t, y=- 16+ (23 sin 6°)t + 155 OC. X=(155 cos 6ºst, y=- 16+(155 cos 6°)t + 23 OD. x = (23 cos 156°)t, y=- 16e +(23 sin 156°yt+6 E x= 155 cos (23°)t, y=- 1612 + 155 sin (23)+6 (b) The ball is in the air for about seconds. (Round to two decimal places as needed.)

Answer 1

solution :- 23 a Хя 1 6 feet Ground devel giren u= ISS feet/sec 0 = 23° Un = ucoso ug = usine = Uxxt = 155 (0523" x & Ans: y = 6+ uyxt- & gt2 = 6+ usinot - 2 x 3 2 2 2 . = 6 + 155.Sin 23'+ -1642 y = - 16t² + 155(Sin 23)2 + 6 And: And (E)

- (b) ball is in the air y = -16+2+155. Sin (73) 4 +6 When all hit the ground y=0 0= -164? + 60.56 + +6 * 16+2-60-567-6-0 t= 3.88 osec. the ball is in air for about 3.88 seconds. - 10 on maxinum heraht maximum height y = -167² +155. sin 23 t + 6 :-16+2 +60.564 +6 for maximum height -324 + 60.56 =0 * 32+ = 60.56 1. 60.56 - 32 seco Sec. :1.89 sec. And

id) maximum maximum height I put t = 21. 89 sec Y max = -16(1.891? +60.56*(1.89) +6 = -57. 1536 + 114.4584 +6 = 63.3048 feet. ce) total horizontal distance flight = horizontal speed & time of - Iss cos 23" x 3.88 feet 553.59 feet and