# 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°

509 Points
11 years ago

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

shubham pansare
15 Points
11 years ago

(a) 30 degrees.

Fawz Naim
37 Points
11 years ago

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

S Sharma
17 Points
4 years ago
Yep correct answer. I'm loving this really specially the questions based upon different types of motion so thanks