A projectile is thrown from a building of height 40m with initial velocity of 20m/s at an angle of 37 degree from horizontal.Find its Time of flight and horizontal range.

Projection velocity is 20 m/s . Direction of velocity makes 37° with horizontal.

 

Horizontal component of velocity,  u_h  = 20 cos37  m/s = 16 m/s

Vertical component of velocity ,     u_v = 20 sin 37   m/s  = 12 m/s

 

To reach the ground , vertical distance to be travelled = 40 m

 

we use the formula " S = u t + (1/2) g t² " to get time.

 

40 = 12 t +(1/2)×9.8×t²   

 

4.9 t² + 12 t - 40 = 0  ...................(1)

 

solution of eqn.(1),  t = (1/2) [ -12 + { 144+(160×4.9) }^1/2 ]  = 9.23 s

 

Horizontal range = u_h × t = 16×9.23 = 148 m