Question

019 10.0 points

Answer 1

Answer:

50.2 m/s

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

$y(t) = h +ut + \frac{1}{2}gt^2$

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:

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

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,

$v_x = \frac{d}{t}=\frac{64}{4.95}=12.9 m/s$

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

$v=\sqrt{v_x^2+v_y^2}=\sqrt{12.9^2+(-48.5)^2}=50.2 m/s$