Answer:0.835 s
Explanation:
Given
Red ball initial velocity([tex]u_r[/tex])=1.1 m/s
height of building(h)=28 m
after 0.5 sec blue ball is thrown with a velocity([tex]u_b[/tex])=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
[tex]h=24.3\times 1.5-\frac{9.81\times 1.5^2}{2}=25.414 m[/tex]
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
[tex]14=1.1\times \left ( t+0.5\right )+\frac{9.81\times \left ( t+0.5\right )^2}{2}-----1[/tex]
for blue ball
[tex]13.1=24.3\times t-\frac{9.81\times t^2}{2}-----2[/tex]
add 1 & 2
we get
[tex]27.1=1.1t+0.55+24.3t+\frac{g\left ( t+0.25\right )}{2}[/tex]
27.1=25.4t+0.55+4.905t+1.226
t=0.835 s
