Question

A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the ground, in units of m/s? Assume that air resistance is negligible. Give the answer as a positive number.

Answer 1

The speed will be 33 m/s.

Step-by-step explanation:

The equations for height and velocity of the ball are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive)

v = velocity at time "t".

Let´s place the origin of the frame of reference on the ground and consider the upward direction as positive.

First, let´s find the time it takes the ball to reach the ground:

y = y0 + v0 · t + 1/2 · g · t²

When the ball hits the ground, y = 0. Then:

0 = 51 m - 7.3 m/s · t - 1/2 · 9.8 m/s² + t²

Solving the quadratic equation:

t = 2.6 s

Now, with this time, we can calculate the velocity at the moment when the ball hits the ground:

v = v0 + g · t

v = -7.3 m/s - 9.8 m/s² · 2.6 s

v = -33 m/s

The speed will be 33 m/s.

Answer 2

32.46m/s

Step-by-step explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

`\frac {Vf^(2)-Vo^2}{2.a} =X`

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

`Vf=√(Vo^2+2gX)\\Vf=√(7.3^2+2(9.81)(51))=32.46m/s`

the speed with the ball hits the ground is 32.46m/s