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VikaD [51]
3 years ago
12

A ball is droped from a height of 16m how much time will pass before the ball hits the ground​

Physics
1 answer:
sergey [27]3 years ago
4 0

Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

h= \frac{1}{2}g{t}^{2}

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

Also, let take

g = 9.8m{s}^{-2}

By substitution we obtain,

16= \frac{1}{2}\times 9.8{t}^{2}

\implies32=9.8{t}^{2}

Diving through by 9.8

\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}

\implies{t}^{2} =3.265

square root both sides, we obtain

\implies t= \sqrt{3.265}

t=1.8s

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3 years ago
You are an engineer in charge of designing a new generation of elevators for a prospective upgrade to the Empire State Building.
DanielleElmas [232]

Answer:

   t_total = 23.757 s

Explanation:

This is a kinematics exercise.

Let's start by calculating the distance and has to reach the limit speed of

v = 18.8 m / s

         v = v₀ + a t₁

the elevator starts with zero speed

         v = a t₁

         t₁ = v / a

         t₁ = 18.8 / 2.40

         t₁ = 7.833 s

in this time he runs

         y₁ = v₀ t₁ + ½ a t₁²

         y₁ = ½ a t₁²

         y₁ = ½ 2.40 7.833²

         y₁ = 73.627 m

This is the time and distance traveled until reaching the maximum speed, which will be constant throughout the rest of the trip.

           x_total = x₁ + x₂

           x₂ = x_total - x₁

           x₂ = 373 - 73,627

           x₂ = 299.373 m

this distance travels at constant speed,

           v = x₂ / t₂

           t₂ = x₂ / v

           t₂ = 299.373 / 18.8

           t₂ = 15.92 s

therefore the total travel time is

           t_total = t₁ + t₂

           t_total = 7.833 + 15.92

           t_total = 23.757 s

6 0
3 years ago
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