Question

A red ball is thrown down with an initial speed of 1.1 m/s from a height of 28 meters above the ground. Then, 0.5 seconds after the red ball is thrown, a blue ball is thrown upward with an initial speed of 24.3 m/s, from a height of 0.9 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s^2. What is the height of the blue ball 2 seconds after the red ball is thrown?

How long after the red ball is thrown are the two balls in the air at the same height?

Answer 1

Answer:0.835 s

Step-by-step explanation:

Given

Red ball initial velocity(
`u_r`)=1.1 m/s

height of building(h)=28 m

after 0.5 sec blue ball is thrown with a velocity(
`u_b`)=24.3 m/s

Height of blue after 2 sec red ball is thrown

i.e. height of blue ball at t=1.5 sec after blue ball is thrown upward

`h=24.3* 1.5-(9.81* 1.5^2)/(2)=25.414 m`

therefore blue ball is at height of 25.414+0.9=26.314 from ground

moment after the two ball is at same height

for red ball

`14=1.1* \left ( t+0.5\right )+(9.81* \left ( t+0.5\right )^2)/(2)-----1`

for blue ball

`13.1=24.3* t-(9.81* t^2)/(2)-----2`

add 1 & 2

we get

`27.1=1.1t+0.55+24.3t+(g\left ( t+0.25\right ))/(2)`

27.1=25.4t+0.55+4.905t+1.226

t=0.835 s