Question

A person throws a ball horizontally from the top of a building that is 24.0 m above the ground level. The ball lands 100 m down range from the base of the building. What was the initial velocity of the ball? What is the speed of the ball when reaches the ground?

Answer 1

a) Initial velocity = 0 m/s

b) Final velocity at ground = 21.7 m/s

Step-by-step explanation:

We consider only vertical motion for the required information.

The ball travels a vertical distance equal to the height of the building. See the attached file diagram.

At point A the initial velocity of the ball will be zero because it just starts from there. So

Initial velocity =Vi= 0 m/s

During its flight the ball travels a vertical distance equal to the height od the building.

Distance =S=24m

Gravitational acceleration = g= 9.8 m/sec2

Velocity at point B = Vf=?

Using

`2gS=Vf^(2)-Vi^(2)`

`==> 2`×
`9.8`×
`24`=
`Vf^(2) -0^(2)`

==>
`470.4=Vf^(2)`

==> Vf =
`√(470.4)`

==> Vf= 21.7 m/sec

So the ball will hit the ground with 21.7 m/s velocity

Answer 2

a) u = 45.21 m/s

b) v = 21.7 m/s

Step-by-step explanation:

Given:

- Initial velocity : u

- Final velocity : v

- Initial height y(0) = 24 m

- Range = 100 m

Find:

- Find the initial and final velocities u and v?

Solution:

- The range of the half projectile motion is given as:

Range = u* sqrt ( 2*y(0) / g)

u = Range* sqrt ( g / 2*y(0))

- Plug in the values and solve for u:

u = 100*sqrt(9.81 / 2*24)

u = 100*0.452078533

u = 45.21 m/s

- Use third equation of motion to calculate the final velocity of the ball when it hits the ground in y - direction:

v^2 - u_y^2 = 2*g*y(0)

- Where vertical component of initial velocity is zero u_y = 0

v^2 = 2*g*y(0)

v = sqrt ( 2*g*y(0))

- plug values in:

v = sqrt ( 2*9.81*24)

v = 21.7 m/s