projectile motion (1 Viewer)

[hasterz]

ex 7.3 4a)

A particle is projected from a point O with speed V and angle of elevation a. At a certain point P on its trajectory, the direction of motion of the particle and the line OP are inclined (in opposite senses) at equal angles b to the horizontal. Show that the time taken to reach P from O is 4Vsina/(3g)

 

[KFunk]

ex 7.3 4a)

A particle is projected from a point O with speed V and angle of elevation a. At a certain point P on its trajectory, the direction of motion of the particle and the line OP are inclined (in opposite senses) at equal angles b to the horizontal. Show that the time taken to reach P from O is 4Vsina/(3g)

The angle that OP makes with the horizontal is defined by the postition (x,y) of the particle such that tanβ = y/x.

The angle of the trajectory (with respect to the horizontal) depends on the velocities V_x and V_y such that tanβ = |V_y /V_x| (absolute value because y will be negative according to the question). This is like when you find the angle of projection based on the initial vertical and horizontal velocities.

V_x = Vcosα
x = Vtcosα

V_y = Vsinα - gt
y = Vtsinα - 1/2.gt²

y/x = |V</sub>y</sub> /V_x| so,

(Vtsin&alpha; - 1/2.gt²)/Vtcos&alpha; = (gt - Vsin&alpha; )/Vcos&alpha;

Vtsin&alpha; - 1/2.gt² = gt² - Vtsin&alpha;

2Vtsin&alpha; = 3/2.gt²

&there4; t = 4Vsin&alpha;/3g