Question

The speed of a projectile when it reaches its maximum height is 0.56 times its speed when it is at half its maximum height. What is the initial projection angle of the projectile?

Answer 1

The initial projection angle of the projectile will be given as ::

Let v_x be the horizontal velocity & v_y be the vertical velocity respectively.

magnitude of speed is given by,   v = v_x² + v_y²    { eq.1 }

At mid-point, v²_{(at mid point)} = v_x² + {v_y (h/2)}²

At top level : v (at top) = (0.56) v_{(at mid point)}                                                           { eq.2 }

we know that, speed at top = horizontal velocity

v_x² = (0.56)² [v_x² + {v_y (h/2)}²]                                                                

solving above eq.

v_x² = (0.3136) v_x² + (0.3136) {v_y (h/2)}²

(0.3136) {v_y (h/2)}² = (1 - 0.3136) v_x²

(0.3136) {v_y (h/2)}² = (0.6864) v_x²

{v_y (h/2)}² = (2.188) v_x²

v_y (h/2) = (2.188) v_x²

v_y (h/2) = (1.479) v_x                                                                                        { eq.3 }

using equation of motion 2,

At the top,     v_y² = v_0y² - 2gh

where, v_y = final vertical velocity = 0 m/s

then v_0y² = 2gh                                                                                          { eq.4 }

At the mid-point,   {v_y (h/2)}² = v_0y² - 2g (h/2)

where, 2g (h/2) = 0

(v_0y² / 2) = {v_y (h/2)}²

v_0y² = 2 {v_y (h/2)}²                                                                                    { eq.5 }

inserting the value of 'v_y(h/2)' in eq.5,

v_0y² = 2 (1.479)² v_x²    

(v_0y / v_x)² = 2 (1.479)²

(v_0y / v_x) = 2 . (1.479)                                                                            { eq.6 }

using a trigonometric identity, = tan^-1 (v_0y / v_x)

= tan^-1 [2 . (1.479)]

= tan^-1 (2.091)

= 64.4 degree