Question

A body is thrown with a horizontal velocity 5m/s from the top of a building of height 12m touches the ground with a velocity of: Please answer........

Answer 1

Answer:

Magnitude of velocity just before touching ground is |⃗v| = 16.13 m/s.

Explanation:

In vertical direction

u₁ = 0

a₁ = g = 9.8 m/s²

s₁ = 12 m

Using the equation of motion for uniform acceleration,

v₁² - u₁² = 2a₁s₁

v₁² - 0² = 2 x 9.8 x 12

v₁² = 235.2

v₁ = 15.34 m/s

In horizontal direction

u₂ = 5 m/s

a₂ = 0

v₂ = u₂ + at

v₂ = 5 + 0*t

v₂ = 5 m/s

Therefore, velocity just before touching ground

⃗v = v₁ i + v₂ j

⃗v = 15.34 i + 5 j

|⃗v| = √(15.34² + 5²)

|⃗v| = 16.13 m/s

Answer 2

Answer:

16.278

Explanation:

DATA:

Vi = 5m/s  (initial velocity)

H = 12m    ( height)

G = 10ms:  (gravity ) constant

Vf = ?         (final velocity)

FORMULA : 2GH = Vf (2) -Vi(2)

     note: (2) represents square

SOLUTION:

 puttting values :

              2 (10) (12) = Vf(2) - 5(2)

              20 * 12      = Vf (2) - 25

              240 +25    = Vf(2)

              265          =$Vf{2}$

Squaring both sides"

               $\sqrt{265}$ = Vf(2)        

                  Vf   = 16.278 m\s