Question

6 An object in shot with a muzzle speed of 175 mph at angle of 60°. It is shot from a au point meters above ground a) Find formula tor the position, *(+). b) How long will it take for the object to hit the ground? c) Find the velocity and the speed of the object when it hits the ground, d) How far doca the object travel horizontally? e) what the maximum height that the object attains?

Answer 1

ry Sol V. = 175 mph. NosinGO V LG Vocoso 15 m H Horizontal Range = x - Vocos boxt = 39.115 X 14.03=548.78m = 175*1.609x5 m/s 18 78.23 m/s position vector ho > ) t = 67.95 + 1 (67.75)² + 4x49x15 67.75 + £ 69.89 2x 4.900 9.8 t= 67.75 +69.89 - 14.03 sec. time taken to reach the ground =t=14.03 sec. x = V. cosbo - 78,23 CosGo' - 39.115 m. vy IsinGO - 9.81€ = 67.75 9.81 x 14.03 = va 39.115 î - 69.88 Ĵ m/s speed = v= Ju? rug? speed - 80.09 m/s Horizontal = 548.78 m :-co-ordinate V. cosboxt = 7823*£xt = 39.115+ Y- cordinate 15tv.singot - £9t=15+78-23x8&t=4x92144 = 15+67.7 - 4.942 xhtyi 39.115t it (15+67.75t-4.94² } when object hits the ground Y-cordinate-o. 15+67.75 -4.9 +²=0 - 49ť - 67.754-1520 80 - 69.884. 119.415€ + 69.84% 80.097 Range

@ Masumum height from ground = 15+ Vo sia zg 78.23)*x Cue 2x9.81x4 15+ 78.23)2 15+ (78.23) ² 3 2 g 248.940. Maximum heeght from ground = H = 248.94m a