Question

2. A projectile is fired from a cannon with initial velocity of 1000 m/s and at an angle of 30° from the horizontal. The point of impact is 1500 m below the point of release. (a) How long does it take for the projectile to hit the ground? (b) Find the projectile's speed at impact. (c) Find the horizontal range of the projectile. (d) Find the maximum height. Neglect air resistance. 1000 m/s 1500 m PREPARED BY: REVIEWED BY: NOTED BY: Engr. DRANREB ERIC D. BERSAMINA Instructor ENGR. JOSEPH RETUMBAN Engr. ALEXA RAY FERNANDO OIC, Mechanical Department College Dean

Answer 1

Solution: u= 1000m/s Given: 30° U= 1000m/s Hį. 1500 m .. . PART (a) Applying equation of motum in y diveetun : - L ol2... a Sy = lyt - gt².. = -1500 = Usun 30 xt -597? => -1500 =1000x sin30x7 -4*9.8x4 = 4.9+2 - 5001 - 1500 to It's 104.957 sec 7 It = 1.75 min 1 time taken to het ground

PART (6) finding projectile speed Again applying equation of motion in ya direction Vy = Uy - gt Vy - 10000 x 5in30 - 9.5x104,957 Į putting value of 't from equation i 4 Vý: - 528.58 m/s 7 Since there is no acceleration in horizontal direction : - So horizontal remains constant throughout projection: At point B. Vy = -528,58 m/s V x = u cosso = 1000 x cosso = 866.025 m/s Net velocity = Jive) +(Vyr N .

Vnet Unet - = (8661.025)*7 6-528.58% 1014.59 m/s. projectile's speed at V impact . Рят с Horizontal range (Sx) . Si į Uxt = u cos 30 t.. LE 1000 x 10530 .104.957 Sx = 90895.42 in 1 Horizontal range of projectie PART (d) Calculation for maximumum height

H2 → Maximum height from refrence - 4 20 (to) ( 4c Applying equation of motion m y - directiom. vý = Uy? - 29 Hz -) ö = (loco x sin 30 ) - 2x9.5 Hz $ At maximum height .H2 = (1000 x sin 30) 2x 9,8 I H2 = 12755010 ml Maximum height from ground 1 Hmax = . H t . Hz 1500 + 12755,10 14.255.1 m