A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown verti

Question

A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?

17 m

21 m

18 m

27 m

31 m

Answer 1

Hi,

distance = height = S m

initial velocity = vi m/s

time = t s

acceleration due to gravity = g = - 9.8 m/s^2 (negative for upward motion)

from the statement of the question

As S = vit + 1/2gt^2

S ball = S stone

note that if time taken by stone is t s then the time taken by ball is t + 2 s

20(t+2) + 1/2*-9.8*(t+2)^2 = 24(t) +1/2*-9.8*t^2

=> 20t + 40 -4.9(t^2 + 4 + 4t) = 24t - 4.9t^2

=> 20t + 40 - 4.9t^2 - 19.6 - 19.6t = 24t - 4.9t^2

=> 20t + 40 -19.6 - 19.6t = 24t

=> t(20 - 19.6 - 24) = -40 + 19.6

=> -23.6t = -20.4

=> t = 20.4/23.6 = 51/59 s.

S = 24t -9.8(t^2) = 24(51/59) - 4.9(51/59)^2 = 17.08 = 17 m

17 m is the correct answer.

Answer 2

Your working equation is

S = VoT - (1/2)gT^2

where

S = distance travelled by the body

Vo = initial velocity

T = travel time

g =  acceleration due to gravity = 9.8 m/sec^2 (constant)

For the ball,

Sb = 20T - (1/2)(9.8)(T^2)

and for the stone,

Ss = 24(T - 2) + (1/2)(9.8)(T - 2)^2

and for them to pass each other from the release point,

Sb = Ss

and therefore

20T - (1/2)(9.8)T^2 = 24(T - 2) - (1/2)(9.8)(T - 2)^2

20T - 4.9T^2 = 24T - 48 - 4.9(T^2 - 4T + 4)

20T - 4.9T^2 = 24T - 48 - 4.9T^2 + 19.6T - 19.6

Combining like terms,

24T + 19.6T - 20T = 48 +19.6

23.6T = 67.6

T = 2.86 sec.

In other words, the stone will pass the ball 2.86 seconds after the ball has been tossed (or 0.86 sec after the stone has been tossed).

Solving for Sb,

Sb = 20(2.86) - (1/2)(9.8)(2.86^2)

Sb = 17.12 meters --- the ball will be passed by the stone at this elevation above the release point. This is the first option in your given choices.