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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