Thread: Hero velocity problem

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 86.0 km/h, to his enemy's car, which is going 103km/h . The enemy's car is 14.5m in front of the hero's when he lets go of the grenade.

If the hero 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.

Find the magnitude of the velocity relative to the earth.

I have no clue how to do this question...it just seems impossible. PLEASE HELP!

let = time (in sec) the grenade is a projectile

position of the enemy car in meters ...



position of the projectile ...



... where is the x-component of grenade's initial velocity

since the idea is for the projectile and enemy car to be at the same place at the same time ...




in the y-direction, the projectile's total displacement is zero ...



... where is the y-component of the grenade's initial velocity.

solving for ...



since the launch angle is , , so .


... substitute for in the equal position equations and solve for

If vy=vx wouldn't vx be 0 then?

Originally Posted by Oldspice1212

If vy=vx wouldn't vx be 0 then?

how did you arrive at that conclusion?

I have no clue what I'm doing . How do you get the relative velocity?!

If the hero 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?

I've set up the necessary equations for you to solve for the magnitude of the grenade's initial velocity. If you cannot follow the proposed solution, then I advise you have a face-to-face consultation with your instructor.

I got it now, thanks :d

Thought I had it, guess not, I got 54.7 km/h?