A stone is projected from the ground with velocity 25 m/s. two seconds later, it just clears a wall 5m high. The angle of projection of the stone is (g=10m/sec2)
(a) 30°
(b) 45°
(c) 50.2°
(d) 60°
let angle of projection is @ then we have
Ux = 25cos@
Uy = 25sin@
after 2 sec it clears a wall of 5m height that means its height from ground is nearly 5m ...
H = Uyt - gt2/2
5 = 25sin@*2 - 10*4/2
5 + 20 = 50sin@
sin@ = 1/2 or @ = 30o
(a) 30 degrees.
intial velocity of the projectile is u=25m/s. time t=2s,h=5m
let angle of projection be theta. the increase in the height of the projectile above the ground is due to the vertical component of the initial velocity which is equal to usintheta. Applying second equation of motion
5=usintheta*2-1/2*10*(2)^2
5=usintheta*2-20
usintheta=25/2
sintheta=25/2*25
sintheta=25/50=1/2
theta=30degree
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