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.

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  1. Your equations are not correct, but I want to thank you for being one of the very few who show their work here. We wish everyone would do that.

    The horizontal distance the grenade travels from the launch point, in earth-based coordinates, is
    V*cos45 + Vj*t

    The distance between cars is
    15.2 + (Ve - Vj)*t = 15.2 + 12.0 t

    The way to get t is NOT to set those distances equal.

    Consider a coordinate system moving with Jones' car. The grenade moves
    V^2/g before hitting the ground when lauched at a 45 degree angle in that coordinate system. The time it takes to hit the ground is
    t = 2Vsin45/g.
    After that time, the separation of cars is
    15.2 + 12 t.
    For the grenade to hit the enemy car,
    V^2/g = 15.2 + 12*sqrt2*V/g

    Solve that quadratic equation for V. Take the posite root.

    B) V is relative to the car in which Jones is travelling. Relative to the Earth, add 43.8 m/s to that.

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  2. Thank for you the explanation!

    But I'd like to ask, how come the last equation isn't

    V^2/g = 15.2 + 12(2Vsin45/g)?

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  3. They are the same, since
    2 sin 45 = sqrt2

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  4. drwls - Your answer isn't correct either. The horizontal distance traveled is V*cos 45*t + Vj*t, and the time it takes for the grenade to reach the ground is SQRT(2Vsin45/g)

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  5. Where does V^2/g come from?

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