# Math Help - Physics with Relative Velocity

1. ## Physics with Relative Velocity

In the latest Indian Jones film, Indy is supposed to throw a grenade from his car, which is going 43.8 m/s, to his enemy's car, which is going 55.8 m/s. The enemy's car is 15.2 m in front of the Indy's when he lets go of the grenade.

A) If Indy 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.

My set up here is
horizontal distance = V * cos45 * Vj * t
[where V is the initial velocity and Vj is Jones' car velocity]

distance between cars = 15.2 + (Vj - Ve) * t
[where Ve is the enemy's car velocity]

I set these two equations equal to each other to find the time it takes the grenade to reach the other car.

(Vcos45)(43.9)(t) = 15.2 + (43.8 - 55.88) * (t)

Then I'm not sure if I'm head in the right direction...

B) Find the magnitude of the velocity relative to the earth.

I'm also not sure how to start this portion.

2. ## Re: Physics with Relative Velocity

Originally Posted by ayyisei
In the latest Indian Jones film, Indy is supposed to throw a grenade from his car, which is going 43.8 m/s, to his enemy's car, which is going 55.8 m/s. The enemy's car is 15.2 m in front of the Indy's when he lets go of the grenade.

A) If Indy 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.

My set up here is
horizontal distance = V * cos45 * Vj * t
[where V is the initial velocity and Vj is Jones' car velocity]

distance between cars = 15.2 + (Vj - Ve) * t
[where Ve is the enemy's car velocity]

I set these two equations equal to each other to find the time it takes the grenade to reach the other car.

(Vcos45)(43.9)(t) = 15.2 + (43.8 - 55.88) * (t)

Then I'm not sure if I'm head in the right direction...

B) Find the magnitude of the velocity relative to the earth.

I'm also not sure how to start this portion.
understand the situation is the same as Indy standing still and the enemy's car moving away from him at 12 m/s with a 15.2 m head start.

let $x = 0$ and $t = 0$ when Indy releases the grenade

time the grenade is in the air ...

$t = \frac{2v_0 \sin(45^\circ)}{g}$

x position for the grenade ...

$x_G = v_0 \cos(45^\circ) t$

x position for the enemy car ...

$x_E = 15.2 + 12t$

sub in the expression for t into the two position expressions, then set the position expressions equal ... solve the resulting quadratic equation for $v_0$ relative to Indy's car

I'll let you think about the last question.

3. ## Re: Physics with Relative Velocity

I found that the initial velocity that Indy throws the grenade is 23.33 and this is the grenade's velocity with respect to Indy's car.

Mastering Physics wants it in km/h so I converted that to 83.88. But I'm actually not sure why it's telling me that both answers are wrong..

For the second part,

I think that to find the magnitude of the grenade with respect to the earth, I'll have to set up a relative velocity equation something like:

$v_G_E = v_G_I + v_I_E$

where

$v_G_E$ is the grenade's velocity with respect to the Earth
$v_G_I$ is the grenade's velocity with respect to Indy's car
$v_I_E$ is Indy's car's velocity with respect to the Earth

I'll have to find the x-component and y-component for $v_G_I$ and $v_I_E$, add them, and find the magnitude from there?

4. ## Re: Physics with Relative Velocity

I get $v_0 = 23.35 \, m/s \approx 84 \, km/hr$

look at the vector sketch to determine the velocity relative to the Earth ...

$v_E = v_0 + v_C$

I get $|v_E| = 62.5 \, m/s$