Projectile Motion: Calculating Initial Velocity for Grenade Throw in Action Film

[limekiwi]

In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 90.0 km/h, to his enemy's car, which is going 127 km/h. The enemy's car is 15.6 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 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.

I've bee reading this problem over and over again, and don't really know how to start. Since I am supposed to find initial velocity, I need the vertical and horizontal component. But from the question, I don't know how to find the horizontal component. Can someone please lead me in the right direction?

 

[AshesToFeonix]

This question didnt make a lot of sense to me... I don't see how it can be solved numerically. The total distance the grenade needs to travel, X final = 15.6 + 37t

to find the velocity, you need the total distance, to get the the total distance you need to know the time it took for the grenade to get from one car to the other. Then again, I'm no rocket scientist, maybe I'm misunderstanding the problem. *shrugs shoulders*

 

[baba89]

Based on the given information, we can use the equations of projectile motion to calculate the initial velocity of the grenade. First, we need to determine the horizontal and vertical components of the initial velocity.

Since the cars are traveling in the same direction and on a level road, we can assume that there is no acceleration in the horizontal direction. Therefore, the horizontal component of the initial velocity will remain constant throughout the motion.

We can use the equation v = d/t to find the horizontal component of the initial velocity. The distance between the two cars is given as 15.6 m and the time it takes for the grenade to reach the other car is the same as the time it takes for the cars to travel that distance. We can assume that the time is the same for both cars since they are traveling at constant speeds. Therefore, the horizontal component of the initial velocity can be calculated as:

v_horizontal = 15.6 m / (90.0 km/h + 127 km/h) = 15.6 m / (217 km/h) = 0.0719 m/s

Next, we can use the equation v = u + at to find the vertical component of the initial velocity. The initial vertical velocity is 0 since the grenade is thrown from the same height as the enemy's car. The acceleration due to gravity is -9.8 m/s^2. The time it takes for the grenade to reach the other car can be calculated using the equation d = ut + 1/2at^2. The distance traveled in the vertical direction is the same as the horizontal distance, 15.6 m. Therefore, we can solve for t:

15.6 m = 0 + 1/2(-9.8 m/s^2)t^2
t = √(15.6 m / 1/2(-9.8 m/s^2))
t = 1.78 s

Now, we can use the equation v = u + at to find the vertical component of the initial velocity:

0 = u_vertical + (-9.8 m/s^2)(1.78 s)
u_vertical = 17.4 m/s

Finally, we can use the Pythagorean theorem to find the magnitude of the initial velocity:

v_initial = √(v_horizontal^2 + u_vertical^2)
v_initial = √(0.0719 m/s)^2 + (17.4