Question

A stone is projected into air at an angle 30° to the horizontal with a speed 20 ms-1 from the top of a tower of height 120 m.
time after which it reaches the ground is (g = 10 ms-2)
(A) 25
(B) 3s
(C) 4s
(D) 65​

Answer 1

Answer:

Given,

∅ = 30°

u = 20 m/s

Height = 120 m

g = 10 m/s²

We need to find the time after which it reaches the ground.

Formula : Time of flight = 2(u sin ∅)/g

⇒ T = 2(20*sin 30°)/10

⇒ T = 2*2*½

⇒ T = 2 seconds

Hence, the after which it reaches the ground is 2 seconds.

Some Important Formulas -

• T = 2(u sin ∅)/g
• R = u² sin2∅/g
• Path of projectile = (tan ∅) x - gx²/2(u cos ∅)²
• H = (u sin ∅)²/2g

Answer 2

Answer:

D)

Explanation:

By resolving the speed

Uy=20sin30=20*1/2=10m/s

Sy=120m

g=10m/s^2

Now,

Sy=Uy+1/2*a*t^2

-120=10t+1/2*(-10)*t^2

-24=2t-t^2

t^2-2t-24=0

Solve to get t=6s

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Hope it helps