Question

019 10.0 points

A ball is thrown horizontally from the top of
a building 120 m high. The ball strikes the
ground 64 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.
020 (ment 1
210,

Answer 1

50.2 m/s

Step-by-step explanation:

First of all, we need to find the time it takes for the ball to reach the ground.

The vertical position of the ball at time t is given by

`y(t) = h +ut + (1)/(2)gt^2`

where

h = 120 m is the initial height

u = 0 is the initial vertical velocity

g = -9.8 m/s^2 is the acceleration of gravity

The ball reaches the ground when y = 0. Substituting into the equation and solving for t, we find the time of light:

`0=h+(1)/(2)gt^2\\t=\sqrt{-(2h)/(g)}=\sqrt{-(2(120))/(-9.8)}=4.95 s`

The vertical component of the velocity of the ball changes following the equation

`v_y(t) = u+gt`

Substituting t = 4.95 s, we find the final vertical velocity of the ball just before reaching the ground:

`v_y=0+(-9.8)(4.95)=-48.5 m/s`

where the negative sign means the direction is downward.

We also can find the horizontal component of the velocity: since we know the horizontal distance travelled is d = 64 m,

`v_x = (d)/(t)=(64)/(4.95)=12.9 m/s`

And the final speed is calculated as the magnitude of the resultant of the two components of the velocity:

`v=√(v_x^2+v_y^2)=√(12.9^2+(-48.5)^2)=50.2 m/s`