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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.

Verified Answer :

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

Further explanation

Acceleration is rate of change of velocity.

\(\large {\boxed {a = \frac{v - u}{t} } }\)

\(\large {\boxed {d = \frac{v + u}{2}~t } }\)

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:

Question A:

First Ball

\(h = H - ut - \frac{1}{2}gt^2\)

\(0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2\)

\(0 = 19.6 - 14.7t - 4.9t^2\)

\(4.9t^2 + 14.7t - 19.6 = 0\)

\(t^2 + 3t - 4 = 0\)

\((t + 4)(t - 1) = 0\)

\((t - 1) = 0\)

\(\boxed {t = 1 ~ second}\)

Second Ball

\(h = H + ut - \frac{1}{2}gt^2\)

\(0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2\)

\(0 = 19.6 + 14.7t - 4.9t^2\)

\(4.9t^2 - 14.7t - 19.6 = 0\)

\(t^2 - 3t - 4 = 0\)

\((t - 4)(t + 1) = 0\)

\((t - 4) = 0\)

\(\boxed {t = 4 ~ seconds}\)

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

\(\Delta t = 4 ~ seconds - 1 ~ second\)

\(\large {\boxed {\Delta t = 3 ~ seconds} }\)

Question B:

First Ball

\(v^2 = u^2 - 2gH\)

\(v^2 = (-14.7)^2 + 2(-9.8)(-19.6)\)

\(v^2 = 600.25\)

\(v = \sqrt {600.25}\)

\(\boxed {v = 24.5 ~ m/s}\)

Second Ball

\(v^2 = u^2 - 2gH\)

\(v^2 = (14.7)^2 + 2(-9.8)(-19.6)\)

\(v^2 = 600.25\)

\(v = \sqrt {600.25}\)

\(\boxed {v = 24.5 ~ m/s}\)

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

Question C:

First Ball

\(h = H - ut - \frac{1}{2}gt^2\)

\(h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2\)

\(\boxed {h = 11.025 ~ m}\)

Second Ball

\(h = H + ut - \frac{1}{2}gt^2\)

\(h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2\)

\(\boxed {h = 25.725 ~ m}\)

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

\(\Delta h = 25.725 ~ m - 11.025 ~ m\)

\(\large {\boxed {\Delta h = 14.7 ~ m} }\)

Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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

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