A basketball player is running at a constant speed of 2.5 m/s when he tosses a basketball upward with a speed of 6.0 m/s. How fa

Answer:

1350 m/s

Clarification:

Bullet speed

The bullet travels 450 m

Sound travels a distance of 450 m

Using the equation S= V × t

==> t= S/V

Thus, the time for the bullet t1=450/vb

and the sound's travel time t2=450/vs

Given that there's a 1/2 sec interval from when the shot is fired to the moment the shooter hears the sound

==>  t1+t2= 1/2 sec

==> 450/vb+450/vs=0.5sec  ---- (1)

Firing sound duration

==> T1 = x/vs

During this time period, the bullet covers 450 m and the sound of impact travels a distance 'x'

The time taken for sound = 450/vb

The time it takes sound to travel distance 'x'= x/vs

therefore let T2= 450/vb + x/vs

However, all this occurs within 3 seconds, i.e., T = 3 sec

because firing takes place before hitting the target, implying the strike sound is heard in time T = T2-T1= 450/vb + x/vs -x/vs

Making T= 3sec

==> 3= 450/vb

==> vb= 1350 m/s

From equations 1 and 2

applying the same principle, in 3 seconds the observer sees the bullet travel 450 m and perceives the sound

Response:

Clarification:

Impulse is equal to change in momentum

mv - mu, where v and u represent the final and initial velocities during the surface impact

For the downward motion of the baseball

v² = u² + 2gh₁

= 2 x 9.8 x 2.25

v = 6.64 m / s

This becomes the initial velocity upon impact.

For the upward movement

v² = u² - 2gh₂

u² = 2 x 9.8 x 1.38

u = 5.2 m / s

This becomes the final velocity post-impact

change in momentum is

m ( final velocity - initial velocity )

.49 ( 5.2 - 6.64 )

=.7056 N.s.

Impulse exerted by the floor in the upward direction is

=.7056 N.s