Question

cu is ground O 4 An object is shot with a muzzle speed of 175 mph at angle of 60°. It is shot from a point meters above a) Find formula for the position, řlt), 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 is the maximum height that the object attains?

Answer 1

% = 1.75mph = 78.23 m/s. Sol vsinGD 160° coso H ism 2 6 y direction. 0,8) ad VC 2S - - 175x1.609 = 281.64 kmph - 281.6445 = 78.23 18 mys. x= V. Cosbot = 78.234 t = 39.115+ Y = 15+ visinot - 9+2 - 15+ 78.23XV2t - 19.81x7 = 15+ 67.750- 4.912 = 39.115 tê t (15+ 67.75€ - 4.9+²)] In asy = ut that? -15 = 67.75 -4.9t2 > 4.9 t² – 67.75t-15=0. t= 67.75 ± √(67.758+4x4-9x15 67.751 69.89 2x4.905 - 2x 4.905 ist 67.75 +69.89 14.03 sec 9.81 x = lo costo = 78.23 cos 60 = 39.115 m/s. Ty : - Vosinbo - 9.81€ = 67.75 9.80 X 14.03 = 69.88 43. ms. velocity = = 39.115 î - 69.884 m/s speed= [ x ² + vy² = {(29-15) + 69.884)2 = 80.086 m/s Speed - Heuting [upward tve) - 80.086 m/s d 548.78 m e x = Vo cos 6oxt 39.1155 14.03 Marimimum Height ~ 25100 29 H - 248.94 m ht = 15+ (78.23) x 3 2x9.81 x 4