Two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m: ds. at the same instant, the other student throws a ball vertically upward at the samr speed. the second ball just misses the balcony on thr way down.

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

Acceleration is rate of change of velocity.

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

Unknown:

A. Δt = ?

B. v = ?

C. Δh = ?

Solution:

The difference in the two ball's time in the air is:

The velocity of each ball as it strikes the ground is 24.5 m/s

The difference in the two ball's height after 0.500 s is:

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

the answer is 12 as i may remember

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