A projectile is fired from the level of the floor whose vertical component is 50 m/s and horizontal component is 40 m/s. Find the horizontal distance covered from the point of launch. Please demonstarte work!
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y = 50 t - 0.5(9.8) t^2
0 = 50 t - 0.5(9.8) t^2
0= t(50 -- 0.5(9.8) t
t= 50/[0.5(9.80]
t = 10.2 sec = time of flight
x = 40t
x=40(10.2)
x = 408 meters = horizontal distance
0 = 50 t - 0.5(9.8) t^2
0= t(50 -- 0.5(9.8) t
t= 50/[0.5(9.80]
t = 10.2 sec = time of flight
x = 40t
x=40(10.2)
x = 408 meters = horizontal distance