Originally Posted by **LittleWing** A projectile is launched upward from the ground with a Velocity of 40m/s at a time of 0 (t=0)..A second projectile is launched at 35m/s upward from the ground 't' seconds later. Both projectiles are at the same height when t = 5..Determine the value of delay time? |

For the first ball, let the height after t = 5 s to be y = h. Using t = 5 s:

y = y0 + v0*t + (1/2)at^2

This equation gives you h.

For the second ball, we are launching in at time T. Again using t = 5 s

y = y0 + v0*(t - T) + (1/2)a(t - T)^2

Using the H from the first equation in the second gives you the time,

-Dan