Question

A projectile launcher sits atop a tall building. The launcher fires the projectile with an initial velocity of 7.0 m/s at an angle of 28° above horizontal. Using the point of launch as a reference point what is the magnitude of the position vector for the projectile 1.5 seconds after being launched?

Answer 1

Vertical component of initial velocity is : v_v = 7 m / s x sin 28^o = 3.286 m / s,

and, the horizontal component is : v_h = 7 m / s x cos 28^o = 6.18 m / s.

Time taken by the projectile to reach at the highest point, where its vertical velocity momentarily becomes zero, be t^/, in seconds.

Hence, v_v - gt^/ = 0

or, t^/ = v_v / g = ( 3.286 / 9.8 ) = 0.335.

Hence, considering a total 1.5 seconds' motion, the downfall time of the projectile was :

t = ( 1.5 - 0.335 ) s = 1.165 s.

If the projectile travelled a vertical distance h, in meters, i this time t, then,

h = 0 x t + gt² / 2 = 9.8 x 1.165² / 2 = 6.65.

Hence, after time 1.5 s, the vertical position of the projectile was : 6.65 m.

Also, horizontal distance travelled by the projectile in this 1.5 s : v_h x 1.5 s = ( 6.18 x 1.5 ) m = 9.27 m.

Hence, the position vector for the projectile 1.5 seconds after being launched is given by :

( 9.27 m , 6.65 m ).