Question

a car falls off a ledge and drops to the ground in 0.9s. if the value of gravitational is 10m/s2,then the speed striking the ground is​

Answer 1

Given :-

• Initial velocity, u = 0 m/s
• Time taken, t = 0.9s
• Acceleration due to gravity, g = 10 m/s²

To find :-

• Final velocity - v

Solution :-

• u = 0 m/s
• t = 0.9s
• g = 10 m/s²

Substituting the value in 1st equation of motion :-

⟹ $\sf v = u + at$

⟹ $\sf v = 0 + 0.9 \times 10$

⟹ $\sf v = 9 m/s$

Velocity with which it strikes the ground is 9 m/s.

Answer 2

Given :-

• Initial Velocity (u) = 0 m/s (as it starts from rest)
• Time Taken (t) = 0.09 s
• Acceleration due gravity (g) = 10 m/s²

To Find :-

• Final Velocity (as when it streaks the ground)

So, in this question we have initial Velocity, time takan & acceleration

So, we can simply get final Velocity

Formula Used :-

• v = u + at

→ v = 0 + 10 × 0.9

→ v = 10 × 9/10

→ v = 1 × 9

→ v = 9

Thus, the final velocity is 9 m/s

So, the speed of car when striking the ground is 9 m/s