Posted by bobjahng2 on .
In the latest Indian Jones film, Indy is supposed to throw a grenade from his car, which is going 79.0 km/h, to his enemy's car, which is going 102 km/h. The enemy's car is 14.3 m in front of the Indy's when he lets go of the grenade.
If Indy throws the grenade so its initial velocity relative to him is at an angle of 45 degrees above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. You can ignore air resistance.
What is the magnitude of velocity compared to earth?

physics 
bobpursley,
horizontal distance=(V*cos45Ic)*time where V is velocity and Ic is Indy's car speed in m/s
distance= 14+(IcEc)*time
set the distances equal
1) 14+(IcEc)t=(Vcos45Ic)t
hf=hi+VsinTheta*t1/2 g t^2
0=Vsin45t4.9 t^2
t= Vsin45/4.9
put that t into 1), and solve for V. This may be challenging algebra. 
physics 
bobjahng2,
uhhhh so i understand everything you did, but i am having issues with the algebra. Any tips/hints?

physics 
Anonymous,
quadratic equation