In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 99.0 km/h , to his enemy's car, which is going 124 km/h . the enemy's car is 16.9 m in front of the hero's when he lets go of the grenade.

First thing to do is to uniform all units needed in the problem. 90 km/h is equal to 25 m/s while 110 km/h is equal to 30.5 m/s. The difference between the two is the reference of the car of the enermy which is equal to 5. 5 m/s.
Since the grenade follows a projectile motion, x or the range of teh shot is equal to v cos theta t where theta is equal to 45 degrees. 
The y-components are expressed as followsy = v sin 45 t - gt^2 where g is 9.8 m/s2v = v sin 45 - gtIn the enemy's side: we wirtex = 15.8 + 5.5 twhere 15.8 is the initial difference in distance between them.

we equate y to zero to actuate the same moment the grenade is thrown upwards0 = v sin 45 t - 9.8 t2v cos45 t = 15.8 + 5.5 tt = 0.144v v by substitution is equal to 17 m/s
x component of this velocity is 17 cos 45  + 25 equal to 37 m/sThe magnitude thus of the velocity by Pythagorean theorem is equal to 39 m/s

Answervery high orbital velocities in a very compact region;