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.

Asked by aviralsrivastavasaksham | 11th Sep, 2019, 09:36: PM

Expert Answer:

Projection velocity is 20 m/s . Direction of velocity makes 37° with horizontal.
 
Horizontal component of velocity,  uh  = 20 cos37  m/s = 16 m/s
Vertical component of velocity ,     uv = 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 = uh × t = 16×9.23 = 148 m

Answered by Thiyagarajan K | 11th Sep, 2019, 10:49: PM

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