What Should the Initial Velocity of the Grenade Be in a High-Speed Car Chase?

[jasonchiang97]

Homework Statement

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 79.0 km/h , to his enemy's car, which is going 125 km/h . The enemy's car is 15.9 m in front of the hero's when he let's go of the grenade.

If the hero throws the grenade so its initial velocity relative to him is at an angle of 45 ∘ 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.

Homework Equations

Vx=Vcos45°*
Vy=Vsin45°
x=-b√(b²-4ac)/2a
d-v_yt+gt²/2
Rx=Vcos45*t

The Attempt at a Solution

villain velocity in m/s=34.72
hero velocity in m/s=21.94
34.72-21.94=8.77m/s

Vx/cos45=V
Vy=(Vx/cos45)sin45
d=(Vx/tan45)t+gt²/2Not sure what to do now.

 

[SteamKing]

Homework Statement

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 79.0 km/h , to his enemy's car, which is going 125 km/h . The enemy's car is 15.9 m in front of the hero's when he let's go of the grenade.

If the hero throws the grenade so its initial velocity relative to him is at an angle of 45 ∘ 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.

Homework Equations

Vx=Vcos45°*
Vy=Vsin45°
x=-b√(b²-4ac)/2a
d-v_yt+gt²/2
Rx=Vcos45*t

The Attempt at a Solution

villain velocity in m/s=34.72
hero velocity in m/s=21.94
34.72-21.94=8.77m/s

Vx/cos45=V
Vy=(Vx/cos45)sin45
d=(Vx/tan45)t+gt²/2Not sure what to do now.

Since the enemy's car is going faster than the hero's, the horizontal velocity of the grenade, relative to the ground, must be greater than the enemy's car in order to catch up with it. Since there is a difference in speed between the two vehicles, the initial separation of 15.9 m between the two cars is going to increase during the time of flight of the grenade.

 

[jasonchiang97]

Since the enemy's car is going faster than the hero's, the horizontal velocity of the grenade, relative to the ground, must be greater than the enemy's car in order to catch up with it. Since there is a difference in speed between the two vehicles, the initial separation of 15.9 m between the two cars is going to increase during the time of flight of the grenade.

That I understand but I'm not quite sure how that information will help me solve it.

 

[SteamKing]

That I understand but I'm not quite sure how that information will help me solve it.

You've figured out the relative velocity between the two cars.
You know what the launch angle of the grenade is.
You know how far apart the cars are initially.
The only thing you are missing is the launch velocity of the grenade.

If nothing else, you can assume a launch velocity and see if the grenade can reach the other car. It may not be an elegant solution, but you should be able to at least bracket an answer in a couple of attempts.

A problem like this is suitable for setting up using a spreadsheet to do the actual calculations.

 

[gneill]

One thing you can do to unclutter the math is take advantage of the launch angle being 45 degrees. That makes the horizontal and vertical components of the grenade velocity (with respect to the hero) equal in magnitude. So just call them both "v". That gets rid of all the trig functions, and you can later combine the components to obtain the magnitude of the velocity vector: $~~~v_o = \sqrt{2 v^2}$.